Chapter 20 · Higher tier

GCSE Maths Pythagoras and Trigonometry (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

Use Pythagoras’ theorem to find the hypotenuse or a shorter side, and to decide whether a triangle is right-angled
Solve Pythagoras problems in context, with coordinates, and in isosceles and equilateral triangles
Label the sides of a right-angled triangle and use sin, cos and tan to find lengths and angles
Solve problems with angles of elevation and depression and with bearings
Know and use the exact values of sin, cos and tan for 0°, 30°, 45°, 60° and 90°
Higher: use the sine rule, the cosine rule and area = ½ab sin C, and solve problems in three dimensions

You should already know

Square numbers and square roots (Chapter 2)
Rounding to decimal places and significant figures (Chapter 4)
Rearranging simple equations and formulae (Chapters 8 and 9)
Angles in a triangle and in an isosceles triangle (Chapter 17)
Area of a triangle (Chapter 18)

Where this leads

Bearings, scale drawings and vectors (Chapter 22)
Trigonometric graphs and functions (Chapter 25, Higher)
Volume and surface area of cones and pyramids (Chapter 19)

Edexcel GCSE (9–1) Mathematics 1MA1, Higher tier. Content references: G6, G20, G21, G22, G23.

Printable worksheets. Every lesson in this chapter has a free worksheet and an answer sheet to download and print.

20.1 Pythagoras’ theorem: finding the hypotenuse and a shorter side: Worksheet (PDF) · Answers (PDF) · lesson page
20.2 Is it right-angled? Pythagorean triples and problems in context: Worksheet (PDF) · Answers (PDF) · lesson page
20.3 Pythagoras with coordinates, isosceles and equilateral triangles: Worksheet (PDF) · Answers (PDF) · lesson page
20.4 Trigonometry: finding sides (SOH CAH TOA): Worksheet (PDF) · Answers (PDF) · lesson page
20.5 Trigonometry: finding angles: Worksheet (PDF) · Answers (PDF) · lesson page
20.6 Angles of elevation and depression, and bearings: Worksheet (PDF) · Answers (PDF) · lesson page
20.7 Exact values of sin, cos and tan: Worksheet (PDF) · Answers (PDF) · lesson page
20.8 The sine rule: Worksheet (PDF) · Answers (PDF) · lesson page
20.9 The cosine rule: Worksheet (PDF) · Answers (PDF) · lesson page
20.10 Area of a triangle: ½ab sin C: Worksheet (PDF) · Answers (PDF) · lesson page
20.11 Trigonometry and Pythagoras in three dimensions, and combined problems: Worksheet (PDF) · Answers (PDF) · lesson page
Check-up and exam-style questions: Worksheet (PDF) · Answers (PDF) · lesson page

Exercise P Do you remember?

Non-calculator

FWork out each of these.

(a)7²

(b)12²

(c)√81

(d)√144

Done

FRound each number to 1 decimal place. (These matter because Pythagoras and trigonometry answers are often rounded.)

(a)7.8102

(b)12.0467

Done

FSolve each equation.

(a)x ÷ 5 = 3.2

(b)12 ÷ x = 3

Done

FThis triangle has a right angle and an angle of 35°. Work out the size of the angle marked x.

Done

FWork out the area of this triangle.

Done

20.1

Pythagoras’ theorem: finding the hypotenuse and a shorter side

Edexcel 1MA1: G20, G6
Grade 3-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Pythagoras’ theorem: finding the hypotenuse and a shorter sideComing soon

A right-angled triangle has one angle of 90° (a square corner). The longest side of a right-angled triangle is the one opposite the right angle. It has its own name: the hypotenuse. The other two sides are called the shorter sides (or the legs).

The hypotenuse is opposite the right angle. It is always the longest side.

More than 2500 years ago, people noticed something special. Draw a square on each side of a right-angled triangle. Then the area of the square on the hypotenuse is exactly equal to the areas of the two other squares added together. In the picture below, the squares on the shorter sides have areas 9 and 16, and 9 + 16 = 25, which is the area of the biggest square.

Squares on the three sides of a 3, 4, 5 triangle: 3² + 4² = 5², that is 9 + 16 = 25.

This is Pythagoras’ theorem. If the shorter sides are a and b and the hypotenuse is c, then

a² + b² = c²

Remember from Chapter 2 that a² means a × a. The theorem only works for right-angled triangles, and c must be the hypotenuse.

Finding the hypotenuse. If you know both shorter sides, square them, add the squares, then take the square root. The square root undoes the squaring, because we want c, not c².

Finding a shorter side. If you know the hypotenuse and one shorter side, square them, subtract (the hypotenuse square is the big one, so you take the smaller square away), then take the square root.

Why add one time and subtract the other? Because the equation is a² + b² = c². Finding c means working out the right-hand side directly. Finding a means moving b² to the other side, which turns the plus into a minus.

A calculator gives the square root of a number that is not a perfect square as a long decimal, so we round. Always read the question to see how accurate to be. The Higher tier also needs answers left as a surd (a square root like √40 simplified to 2√10) when the question asks for an exact answer.

Key ideas

Hypotenuse = the side opposite the right angle = the longest side.
Pythagoras: a² + b² = c² (c is the hypotenuse).
Finding the hypotenuse: add the squares, then square root.
Finding a shorter side: subtract the squares, then square root.
Check: the hypotenuse must be longer than both shorter sides.

Worked example 20.1.1

Work out the length of the hypotenuse, x.

Key words. The hypotenuse is the longest side of a right-angled triangle; it is always opposite the right angle. Squaring a number means multiplying it by itself. The square root (√) undoes squaring.

Method. To find the hypotenuse, square the two shorter sides, add them, then take the square root.

Spot the hypotenuse. The hypotenuse is the side opposite the right angle (the little square) and it is the longest side. The side we want, x, is the one opposite the right angle, so it is the hypotenuse. We know the two shorter sides.
Shorter sides: 5 cm and 12 cm.
Write down Pythagoras’ theorem and put the numbers in. For a right-angled triangle, (shorter side)² + (shorter side)² = (hypotenuse)².
a² + b² = c²
5² + 12² = x²
Square each short side. Squaring means multiplying the number by itself.
5² = 5 × 5 = 25
12² = 12 × 12 = 144
Add the two squares. This gives x², the square of the hypotenuse.
x² = 25 + 144 = 169
Take the square root. We want x, not x². The square root undoes the squaring.
x = √169 = 13 cm

Check. The hypotenuse must be the longest side: 13 is bigger than 5 and 12. Also 5² + 12² = 169 = 13². ✓

Watch out. For the hypotenuse you add the squares. Do not stop at x²: the last step is the square root.

Worked example 20.1.2

Work out the length of the hypotenuse, x. Give your answer correct to 1 decimal place.

Method. To find the hypotenuse, square the two shorter sides, add them, then take the square root.

Spot the hypotenuse. The hypotenuse is the side opposite the right angle (the little square) and it is the longest side. The side we want, x, is the one opposite the right angle, so it is the hypotenuse. We know the two shorter sides.
Shorter sides: 7 cm and 9 cm.
Write down Pythagoras’ theorem and put the numbers in. For a right-angled triangle, (shorter side)² + (shorter side)² = (hypotenuse)².
a² + b² = c²
7² + 9² = x²
Square each short side. Squaring means multiplying the number by itself.
7² = 7 × 7 = 49
9² = 9 × 9 = 81
Add the two squares. This gives x², the square of the hypotenuse.
x² = 49 + 81 = 130
Take the square root. We want x, not x². The square root undoes the squaring. Use the √ button on the calculator.
x = √130 = 11.40175……
Round to 1 d.p.. The question tells us how accurate to be.
x = 11.4 cm

Check. Square the answer: 11.4² = 129.96, which is close to 130. The hypotenuse 11.4 is longer than both 7 and 9. ✓

Watch out. For the hypotenuse you add the squares. Do not stop at x²: the last step is the square root.

Worked example 20.1.3

Work out the length of the shorter side, x.

Method. To find a shorter side, square the hypotenuse and the known side, subtract, then take the square root.

Spot the hypotenuse. The hypotenuse is opposite the right angle and is the longest side. This time it is given, and we are looking for one of the two shorter sides.
Hypotenuse = 17 cm. The other known side = 8 cm. The side we want is x.
Write down Pythagoras’ theorem and put the numbers in. The hypotenuse is always the one on its own on the right-hand side: (shorter)² + (shorter)² = (hypotenuse)².
a² + b² = c²
x² + 8² = 17²
Square the known numbers. Squaring means multiplying the number by itself.
8² = 8 × 8 = 64
17² = 17 × 17 = 289
x² + 64 = 289
Subtract 64 from both sides. We want x² on its own. Taking 64 away from both sides keeps the equation balanced.
x² = 289 − 64 = 225
Take the square root. We want x, not x².
x = √225 = 15 cm

Check. Put it back: 15² + 8² = 225 + 64 = 289 = 17². ✓ And 15 is shorter than the hypotenuse 17.

Watch out. When you are finding a shorter side you subtract. A shorter side must always be smaller than the hypotenuse; if your answer is bigger, you added by mistake.

Worked example 20.1.4

Work out the length x. Give your answer correct to 1 decimal place.

Method. To find a shorter side, square the hypotenuse and the known side, subtract, then take the square root.

Spot the hypotenuse. The hypotenuse is opposite the right angle and is the longest side. This time it is given, and we are looking for one of the two shorter sides.
Hypotenuse = 9.4 m. The other known side = 6.2 m. The side we want is x.
Write down Pythagoras’ theorem and put the numbers in. The hypotenuse is always the one on its own on the right-hand side: (shorter)² + (shorter)² = (hypotenuse)².
a² + b² = c²
x² + 6.2² = 9.4²
Square the known numbers. Squaring means multiplying the number by itself.
6.2² = 6.2 × 6.2 = 38.44
9.4² = 9.4 × 9.4 = 88.36
x² + 38.44 = 88.36
Subtract 38.44 from both sides. We want x² on its own. Taking 38.44 away from both sides keeps the equation balanced.
x² = 88.36 − 38.44 = 49.92
Take the square root. We want x, not x².
x = √49.92 = 7.06540…
Round to 1 d.p.. The question tells us how accurate to be.
x = 7.1 m

Check. Put it back: 7.1² + 6.2² = 50.41 + 38.44 = 88.85, which is close to 9.4² = 88.36. ✓ The answer is shorter than the hypotenuse 9.4.

Watch out. When you are finding a shorter side you subtract. A shorter side must always be smaller than the hypotenuse; if your answer is bigger, you added by mistake.

Worked example 20.1.5

Work out the length x. Give your answer as a simplified surd.

Method. To find the hypotenuse, square the two shorter sides, add them, then take the square root.

Spot the hypotenuse. The hypotenuse is the side opposite the right angle (the little square) and it is the longest side. The side we want, x, is the one opposite the right angle, so it is the hypotenuse. We know the two shorter sides.
Shorter sides: 6 cm and 10 cm.
Write down Pythagoras’ theorem and put the numbers in. For a right-angled triangle, (shorter side)² + (shorter side)² = (hypotenuse)².
a² + b² = c²
6² + 10² = x²
Square each short side. Squaring means multiplying the number by itself.
6² = 6 × 6 = 36
10² = 10 × 10 = 100
Add the two squares. This gives x², the square of the hypotenuse.
x² = 36 + 100 = 136
Take the square root and leave it exact. The question wants an exact answer, so we do not use a decimal. We simplify the surd instead.
x = √136
Find the biggest square number that divides 136: 136 = 4 × 34
√136 = √4 × √34 = 2√34
x = 2√34 cm

Check. On a calculator 2√34 = 11.66190… and √136 = 11.66190…. It is longer than 6 and 10. ✓

Watch out. For the hypotenuse you add the squares. Do not stop at x²: the last step is the square root.

Watch out

Do not add the two shorter sides and stop, and do not forget the last step. If c² = 169 the answer is c = 13, not 169. Also, always check that the answer for a shorter side is smaller than the hypotenuse. If it is bigger you added when you should have subtracted.

20.2

Is it right-angled? Pythagorean triples and problems in context

Edexcel 1MA1: G20, G6
Grade 4-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Is it right-angled? Pythagorean triples and problems in contextComing soon

Pythagoras’ theorem works backwards too. If the squares of the two shorter sides add up to the square of the longest side, then the triangle must have a right angle. If they do not, it has no right angle. This gives us a test for a right angle using only lengths.

Test: 7² + 24² = 49 + 576 = 625 and 25² = 625. They are equal, so the triangle has a right angle opposite the 25 cm side.

Whole numbers that fit the theorem are called a Pythagorean triple. The most famous is 3, 4, 5 (9 + 16 = 25). Others you should recognise are 5, 12, 13 and 8, 15, 17 and 7, 24, 25. If you multiply every number in a triple by the same amount you get another triple, so 6, 8, 10 and 30, 40, 50 also work. Builders use 3, 4, 5 to check that a corner is square: they measure 3 m along one wall and 4 m along the other and see if the gap between the ends is exactly 5 m.

If the sum of the squares of the two shorter sides is less than the square of the longest side, the biggest angle is more than 90° (obtuse). If it is more, the biggest angle is less than 90°. You do not need this to answer most questions, but it explains why the test says “not right-angled”.

Problems in context. Many real problems hide a right-angled triangle: a ladder against a wall, the diagonal of a rectangle, a ship that sails north then east, a wire holding up a pole. Follow these steps each time:

Draw a sketch and mark the right angle.
Mark the lengths you know, and call the one you want x.
Decide whether x is the hypotenuse (the longest side, opposite the right angle) or a shorter side.
Use Pythagoras: add the squares for the hypotenuse, subtract for a shorter side.
Check that the answer is sensible, then write it with its unit.

Key ideas

Test for a right angle: (short)² + (short)² = (longest)².
A Pythagorean triple is three whole numbers that fit: 3,4,5; 5,12,13; 8,15,17; 7,24,25.
Multiplying a triple by the same number gives another triple.
Always draw a sketch for a word problem and mark the right angle.

Worked example 20.2.1

A triangle has sides of length 7 cm, 24 cm and 25 cm. Is it a right-angled triangle? You must show your working.

Method. Square all three sides. If (short)² + (short)² equals (longest)², the triangle has a right angle.

Find the longest side. If the triangle is right-angled, the longest side must be the hypotenuse, so it goes on its own on one side of the equation.
Sides in order: 7, 24, 25. Longest = 25 cm.
Square the two shorter sides and add. This is the left-hand side of Pythagoras’ theorem.
7² + 24² = 49 + 576 = 625
Square the longest side. This is the right-hand side.
25² = 625
Compare the two answers. If they are equal, Pythagoras’ theorem works, so the triangle is right-angled. If not, it is not.
625 = 625. They are equal, so the triangle is right-angled.

Check. Sides that fit Pythagoras’ theorem are called a Pythagorean triple. This triple cannot be made smaller, so it is a basic one. ✓

Watch out. Add the squares of the two shorter sides and compare with the square of the longest side. Do not add the longest side by mistake.

Worked example 20.2.2

A triangle has sides of length 6 cm, 7 cm and 10 cm. Is it a right-angled triangle? You must show your working.

Method. Square all three sides. If (short)² + (short)² equals (longest)², the triangle has a right angle.

Find the longest side. If the triangle is right-angled, the longest side must be the hypotenuse, so it goes on its own on one side of the equation.
Sides in order: 6, 7, 10. Longest = 10 cm.
Square the two shorter sides and add. This is the left-hand side of Pythagoras’ theorem.
6² + 7² = 36 + 49 = 85
Square the longest side. This is the right-hand side.
10² = 100
Compare the two answers. If they are equal, Pythagoras’ theorem works, so the triangle is right-angled. If not, it is not.
85 ≠ 100. They are not equal, so the triangle is not right-angled.

Check. If it were right-angled the longest side would be √(36 + 49) = 9.21954…, but it is 10. ✓ The longest side is longer than that, so the biggest angle is more than 90°.

Watch out. Add the squares of the two shorter sides and compare with the square of the longest side. Do not add the longest side by mistake.

Worked example 20.2.3

A ladder 5.6 m long leans against a vertical wall. The foot of the ladder is 1.4 m from the base of the wall, on horizontal ground. How far up the wall does the ladder reach? Give your answer correct to 2 decimal places.

Method. To find a shorter side, square the hypotenuse and the known side, subtract, then take the square root.

Draw a sketch. The wall is vertical and the ground is horizontal, so they make a right angle. The ladder is the hypotenuse because it is opposite that right angle.
Ladder = 5.6 m (hypotenuse). Foot of ladder to the wall = 1.4 m. Height up the wall = h.
Spot the hypotenuse. The hypotenuse is opposite the right angle and is the longest side. This time it is given, and we are looking for one of the two shorter sides.
Hypotenuse = 5.6 m. The other known side = 1.4 m. The side we want is h.
Write down Pythagoras’ theorem and put the numbers in. The hypotenuse is always the one on its own on the right-hand side: (shorter)² + (shorter)² = (hypotenuse)².
a² + b² = c²
h² + 1.4² = 5.6²
Square the known numbers. Squaring means multiplying the number by itself.
1.4² = 1.4 × 1.4 = 1.96
5.6² = 5.6 × 5.6 = 31.36
h² + 1.96 = 31.36
Subtract 1.96 from both sides. We want h² on its own. Taking 1.96 away from both sides keeps the equation balanced.
h² = 31.36 − 1.96 = 29.4
Take the square root. We want h, not h².
h = √29.4 = 5.42217…
Round to 2 d.p.. The question tells us how accurate to be.
h = 5.42 m

Check. Put it back: 5.42² + 1.4² = 29.38 + 1.96 = 31.34, which is close to 5.6² = 31.36. ✓ The answer is shorter than the hypotenuse 5.6.

Watch out. When you are finding a shorter side you subtract. A shorter side must always be smaller than the hypotenuse; if your answer is bigger, you added by mistake.

Worked example 20.2.4

For any whole numbers m and n with m > n, show that m² − n², 2mn and m² + n² are the sides of a right-angled triangle. Then use m = 3 and n = 2 to make a triple.

Key words. A Pythagorean triple is a set of three whole numbers a, b, c with a² + b² = c². To show that means to give every step of a proof.

Method. The longest side should be m² + n². Show that the two shorter sides squared and added equal its square, using algebra.

Square the first short side. Use (p − q)² = p² − 2pq + q² with p = m² and q = n².
(m² − n²)² = m⁴ − 2m²n² + n⁴
Square the second short side. Square both the 2 and the mn.
(2mn)² = 4m²n²
Add the two squares. Collect like terms: −2m²n² + 4m²n² = 2m²n².
m⁴ − 2m²n² + n⁴ + 4m²n² = m⁴ + 2m²n² + n⁴
Square the longest side. Use (p + q)² = p² + 2pq + q².
(m² + n²)² = m⁴ + 2m²n² + n⁴
Compare. The two results are the same, so Pythagoras’ theorem holds and the triangle has a right angle.
(m² − n²)² + (2mn)² = (m² + n²)²
Make a triple with m = 3 and n = 2. Substitute into the three expressions.
3² − 2² = 9 − 4 = 5
2 × 3 × 2 = 12
3² + 2² = 9 + 4 = 13
The triple is 5, 12, 13.

Check. 5² + 12² = 25 + 144 = 169 = 13². ✓

Watch out. Expand brackets carefully: (m² − n²)² is not m⁴ − n⁴.

Watch out

When testing for a right angle, put the longest side on its own. A common mistake is to add the square of the longest side to another one. Also, “close enough” is not good enough: 6, 7, 10 has 36 + 49 = 85, which is not 100, so there is no right angle.

20.3

Pythagoras with coordinates, isosceles and equilateral triangles

Edexcel 1MA1: G20, G6, G11
Grade 4-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Pythagoras with coordinates, isosceles and equilateral trianglesComing soon

Distance between two points. To find the distance between two points on a grid, imagine a right-angled triangle between them. The straight line joining the points is the hypotenuse. The horizontal step is the difference of the x-coordinates and the vertical step is the difference of the y-coordinates. Then use Pythagoras. A minus sign in a step does not matter, because squaring turns it into a positive number.

A(1, 2) and B(7, 10): the steps are 7 − 1 = 6 across and 10 − 2 = 8 up, so AB² = 6² + 8² = 100 and AB = 10.

Isosceles triangles. An isosceles triangle has two equal sides, and the angles at the base are equal. If you draw the line of symmetry from the top down to the base, it hits the base at 90° and cuts it exactly in half. That makes two right-angled triangles that are mirror images of each other. Their hypotenuse is one of the equal sides, one shorter side is half the base and the other shorter side is the height of the triangle. Once you know the height you can find the area: ½ × base × height.

The height splits the base into two halves of 5 cm. In each right-angled triangle: h² + 5² = 13².

Equilateral triangles. An equilateral triangle has three equal sides and three 60° angles. It is a special isosceles triangle, so the same trick works. For a side of length s the half-base is s ÷ 2, so the height h satisfies h² + (s ÷ 2)² = s².

Equilateral triangle of side 8 cm: h² + 4² = 8².

Higher tier: you may be asked for an exact answer. Do not round. Leave square roots in the answer and simplify them, for example √52 = √4 × √13 = 2√13.

Key ideas

Distance between points: (x step)² + (y step)² = distance².
An isosceles triangle splits into two right-angled triangles: use half the base.
Area of the triangle = ½ × whole base × height.
Equilateral triangle: height² = side² − (½ side)².

Worked example 20.3.1

Work out the distance between the points A(1, 2) and B(7, 10).

Method. The distance between two points is the hypotenuse of a right-angled triangle whose shorter sides are the horizontal and vertical steps between them.

Draw a right-angled triangle between the two points. The line from one point to the other is the hypotenuse. The two shorter sides are the horizontal step and the vertical step.
Horizontal step across, vertical step up or down.
Find the horizontal and vertical steps. Subtract the x-coordinates, then subtract the y-coordinates. Only the size of each step matters, so ignore any minus sign.
Horizontal: 7 − 1 = 6, so the step is 6
Vertical: 10 − 2 = 8, so the step is 8
Use Pythagoras’ theorem. The distance d is the hypotenuse.
d² = 6² + 8² = 36 + 64 = 100
Take the square root. We want d, not d².
d = √100 = 10

Check. 6, 8, 10 fits a Pythagorean triple pattern: 36 + 64 = 100. ✓

Watch out. Subtract the coordinates in the same order for x and y. A negative step is fine: squaring removes the minus sign.

Worked example 20.3.2

The points P(−3, 4) and Q(5, −2) are joined by a straight line. Work out the length of PQ.

Method. The distance between two points is the hypotenuse of a right-angled triangle whose shorter sides are the horizontal and vertical steps between them.

Draw a right-angled triangle between the two points. The line from one point to the other is the hypotenuse. The two shorter sides are the horizontal step and the vertical step.
Horizontal step across, vertical step up or down.
Find the horizontal and vertical steps. Subtract the x-coordinates, then subtract the y-coordinates. Only the size of each step matters, so ignore any minus sign.
Horizontal: 5 − (−3) = 8, so the step is 8
Vertical: −2 − 4 = −6, so the step is 6
Use Pythagoras’ theorem. The distance d is the hypotenuse.
d² = 8² + 6² = 64 + 36 = 100
Take the square root. We want d, not d².
d = √100 = 10

Check. 8, 6, 10 fits a Pythagorean triple pattern: 64 + 36 = 100. ✓

Watch out. Subtract the coordinates in the same order for x and y. A negative step is fine: squaring removes the minus sign.

Worked example 20.3.3

An isosceles triangle has two sides of length 13 cm and a base of 10 cm. Work out the area of the triangle.

Method. Split the isosceles triangle down the middle into two right-angled triangles, then use Pythagoras’ theorem.

Draw the line of symmetry from the top to the middle of the base. In an isosceles triangle this line is the perpendicular height. It cuts the base in half and splits the triangle into two identical right-angled triangles.
Half the base = 10 ÷ 2 = 5 cm
Pick one right-angled triangle. Its hypotenuse is the slanting side, 13 cm. Its shorter sides are the half-base and the height h.
h² + 5² = 13²
Square and subtract. We are finding a shorter side, so we subtract.
h² = 169 − 25 = 144
Take the square root. The height is h, not h².
h = √144 = 12 cm
Use area of a triangle = ½ × base × height. The base is the whole base, not the half. Use the exact height from the calculator to avoid rounding errors.
Area = ½ × 10 × 12 = 60
Write the area with its units. Area is measured in square units. The answer is exact, so no rounding is needed.
Area = 60 cm²

Check. Another way: two small triangles each have area ½ × 5 × 12 = 30.00; doubling gives 60.00, matching 60. ✓

Watch out. Use half the base in Pythagoras’ theorem, but the whole base in the area formula.

Worked example 20.3.4

An equilateral triangle has sides of length 8 cm. Work out its height, correct to 2 decimal places.

Method. Split the isosceles triangle down the middle into two right-angled triangles, then use Pythagoras’ theorem.

Draw the line of symmetry from the top to the middle of the base. In an isosceles triangle this line is the perpendicular height. It cuts the base in half and splits the triangle into two identical right-angled triangles.
Half the base = 8 ÷ 2 = 4 cm
Pick one right-angled triangle. Its hypotenuse is the slanting side, 8 cm. Its shorter sides are the half-base and the height h.
h² + 4² = 8²
Square and subtract. We are finding a shorter side, so we subtract.
h² = 64 − 16 = 48
Take the square root. The height is h, not h².
h = √48 = 6.92820…
h = 6.93 cm

Check. The height (6.93) must be shorter than the slanting side (8). Also 6.93² + 4² = 64.02, close to 64. ✓

Watch out. Use half the base in Pythagoras’ theorem, but the whole base in the area formula.

Worked example 20.3.5

A is the point (−2, 1) and B is the point (4, 5). Show that the length of AB is 2√13. Give your answer as a simplified surd.

Method. The distance between two points is the hypotenuse of a right-angled triangle whose shorter sides are the horizontal and vertical steps between them.

Draw a right-angled triangle between the two points. The line from one point to the other is the hypotenuse. The two shorter sides are the horizontal step and the vertical step.
Horizontal step across, vertical step up or down.
Find the horizontal and vertical steps. Subtract the x-coordinates, then subtract the y-coordinates. Only the size of each step matters, so ignore any minus sign.
Horizontal: 4 − (−2) = 6, so the step is 6
Vertical: 5 − 1 = 4, so the step is 4
Use Pythagoras’ theorem. The distance d is the hypotenuse.
d² = 6² + 4² = 36 + 16 = 52
Take the square root and leave it exact. An exact answer is wanted, so we simplify the surd.
d = √52
Find the biggest square number that divides 52: 52 = 4 × 13
√52 = √4 × √13 = 2√13

Check. 2√13 = 7.21110… on a calculator, and the line looks about that long on the grid. ✓

Watch out. Subtract the coordinates in the same order for x and y. A negative step is fine: squaring removes the minus sign.

Worked example 20.3.6

An equilateral triangle has side 2a. Show that its height is a√3, and find its area in terms of a.

Method. Split the triangle in half down its line of symmetry. We get a right-angled triangle with hypotenuse 2a, shorter sides a and h.

Write Pythagoras for the half-triangle. The hypotenuse is the full side 2a. The base of the half-triangle is half of 2a, which is a.
h² + a² = (2a)²
Expand and rearrange. (2a)² = 4a². Subtract a² from both sides.
h² = 4a² − a² = 3a²
Take the square root. √(3a²) = √3 × √(a²) = √3 × a, because a is a positive length.
h = a√3
Find the area. Area = ½ × whole base × height.
Area = ½ × 2a × a√3 = a²√3

Check. Test with a = 4: side 8, height = 4√3 = 6.93 (matches the earlier example). ✓

Watch out. The base of the half-triangle is a (half of 2a), not 2a.

Watch out

For an isosceles triangle, use half the base in Pythagoras’ theorem, but the whole base in the area formula ½ × base × height. Mixing these up is the most common mistake.

20.4

Trigonometry: finding sides (SOH CAH TOA)

Edexcel 1MA1: G20
Grade 4-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Trigonometry: finding sides (SOH CAH TOA)Coming soon

Pythagoras uses two sides to find the third. But what if you know only one side and one angle? That is where trigonometry helps. It links the angles of a right-angled triangle to the ratios of its sides.

Look at the picture. All three triangles have a 30° angle and a right angle, so they are similar (the same shape, different sizes). Work out opposite ÷ hypotenuse for each: 1 ÷ 2, 2 ÷ 4 and 3 ÷ 6. They all equal 0.5. This is no accident. In any right-angled triangle with a 30° angle, the ratio of the side opposite the angle to the hypotenuse is always 0.5. We call this ratio the sine of the angle, so sin 30° = 0.5.

Three similar right-angled triangles with a 30° angle. Opposite (blue) ÷ hypotenuse (green): 1÷2, 2÷4, 3÷6 are all 0.5.

Naming the sides. We name the sides from the point of view of the angle we are using (call it θ, said “theta”).

The hypotenuse (H) is opposite the right angle. It is the longest side and it never changes.
The opposite (O) is across the triangle from θ. It does not touch θ.
The adjacent (A) is next to θ. “Adjacent” means “next to”. It is the side left over that is not the hypotenuse.

The names of the sides depend on which angle θ you stand at.

There are three trigonometric ratios. Each compares two sides.

sin θ = OH cos θ = AH tan θ = OA

A memory aid: SOH CAH TOA. Sin = Opposite over Hypotenuse, Cos = Adjacent over Hypotenuse, Tan = Opposite over Adjacent. You can also use the triangle trick below: cover the one you want and read what is left.

Cover up the letter you want. If you cover S you see O over H, so sin = O ÷ H. If you cover O you see S × H, so O = sin × H.

The method for finding a side.

Label the sides O, A, H from the angle θ.
Choose the ratio: look at the side you know and the side you want, and find the two letters in SOH CAH TOA.
Write the ratio with the numbers: sin θ = O ÷ H becomes, for example, sin 35° = x ÷ 8.2.
Rearrange to get x alone. If x is on the top, multiply both sides by the number on the bottom. If x is on the bottom, you will multiply by x and then divide.
Calculate with your calculator in degree mode and round sensibly.

The shortcut for step 4: if the unknown is on the top you multiply, and if the unknown is on the bottom you divide. Use this as a check but know why it works. You can also check that your answer is sensible. The hypotenuse must be the longest side.

Key ideas

Label O, A, H from the angle θ. H is opposite the right angle.
SOH: sin θ = O ÷ H. CAH: cos θ = A ÷ H. TOA: tan θ = O ÷ A.
Choose the ratio from the two sides involved (one known, one wanted).
Unknown on top: multiply. Unknown on the bottom: divide.
Calculator in degree mode.

Worked example 20.4.1

Work out the length x. Give your answer correct to 1 decimal place.

Key words. In a right-angled triangle the hypotenuse is the longest side (opposite the right angle). Measured from the angle θ: the opposite side is directly across from θ, and the adjacent side is next to θ (it is not the hypotenuse). A trigonometric ratio compares two of the sides.

Method. Label the sides, pick SOH, CAH or TOA, write the ratio with the numbers, then rearrange to get the unknown side.

Label the sides. Stand at the angle θ. The hypotenuse is opposite the right angle. The opposite is across from θ and the adjacent is next to θ.
8.2 cm is the hypotenuse.
x is the opposite.
Choose the ratio. We use the ratio that contains the two sides involved: the hypotenuse and the opposite. Use SOH CAH TOA to pick it.
SOH stands for Sin, Opposite, Hypotenuse:
sin θ = OH (O = opposite, H = hypotenuse)
Put the numbers into the ratio. Replace θ with the angle and the sides with the lengths.
sin 35° = x8.2
Multiply both sides by 8.2. x is being divided by 8.2. Multiplying both sides by 8.2 gets x on its own.
x = 8.2 × sin 35°
Use the calculator (degree mode). First find sin 35°, then finish the calculation.
sin 35° = 0.57357…
x = 8.2 × 0.57357… = 4.70332…
Round to 1 d.p.. Give the answer to the accuracy asked, with the unit.
x = 4.7 cm

Check. The other acute angle is 90° − 35° = 55°. From that angle the opposite and adjacent sides swap, so we use cos: x = 8.2 × cos 55° = 4.70332…, the same answer. ✓

Watch out. Check your calculator is in degree mode (DEG). The unknown is on the top, so you multiply. A quick check: the hypotenuse must be the longest side.

Worked example 20.4.2

Work out the length x. Give your answer correct to 1 decimal place.

Method. Label the sides, pick SOH, CAH or TOA, write the ratio with the numbers, then rearrange to get the unknown side.

Label the sides. Stand at the angle θ. The hypotenuse is opposite the right angle. The opposite is across from θ and the adjacent is next to θ.
12 m is the adjacent.
x is the hypotenuse.
Choose the ratio. We use the ratio that contains the two sides involved: the adjacent and the hypotenuse. Use SOH CAH TOA to pick it.
CAH stands for Cos, Adjacent, Hypotenuse:
cos θ = AH (A = adjacent, H = hypotenuse)
Put the numbers into the ratio. Replace θ with the angle and the sides with the lengths.
cos 40° = 12x
Multiply both sides by x. x is on the bottom of the fraction. Multiplying by x brings it to the top.
x × cos 40° = 12
Divide both sides by cos 40°. This leaves x on its own.
x = 12cos 40° = 12 ÷ cos 40°
Use the calculator (degree mode). First find cos 40°, then finish the calculation.
cos 40° = 0.76604…
x = 12 ÷ 0.76604… = 15.66488…
Round to 1 d.p.. Give the answer to the accuracy asked, with the unit.
x = 15.7 m

Check. The other acute angle is 90° − 40° = 50°. From that angle the opposite and adjacent sides swap, so we use sin: x = 12 ÷ sin 50° = 15.66488…, the same answer. ✓

Watch out. Check your calculator is in degree mode (DEG). The unknown is on the bottom, so you divide. A quick check: the hypotenuse must be the longest side.

Worked example 20.4.3

Work out the length x. Give your answer correct to 1 decimal place.

Method. Label the sides, pick SOH, CAH or TOA, write the ratio with the numbers, then rearrange to get the unknown side.

Label the sides. Stand at the angle θ. The hypotenuse is opposite the right angle. The opposite is across from θ and the adjacent is next to θ.
7.5 cm is the adjacent.
x is the opposite.
Choose the ratio. We use the ratio that contains the two sides involved: the adjacent and the opposite. Use SOH CAH TOA to pick it.
TOA stands for Tan, Opposite, Adjacent:
tan θ = OA (O = opposite, A = adjacent)
Put the numbers into the ratio. Replace θ with the angle and the sides with the lengths.
tan 52° = x7.5
Multiply both sides by 7.5. x is being divided by 7.5. Multiplying both sides by 7.5 gets x on its own.
x = 7.5 × tan 52°
Use the calculator (degree mode). First find tan 52°, then finish the calculation.
tan 52° = 1.27994…
x = 7.5 × 1.27994… = 9.59956…
Round to 1 d.p.. Give the answer to the accuracy asked, with the unit.
x = 9.6 cm

Check. The other acute angle is 90° − 52° = 38°. From that angle the opposite and adjacent sides swap, so we use tan: x = 7.5 ÷ tan 38° = 9.59956…, the same answer. ✓

Watch out. Check your calculator is in degree mode (DEG). The unknown is on the top, so you multiply. A quick check: the hypotenuse must be the longest side.

Worked example 20.4.4

Work out the length x. Give your answer correct to 3 significant figures.

Method. Label the sides, pick SOH, CAH or TOA, write the ratio with the numbers, then rearrange to get the unknown side.

Label the sides. Stand at the angle θ. The hypotenuse is opposite the right angle. The opposite is across from θ and the adjacent is next to θ.
14.2 mm is the opposite.
x is the adjacent.
Choose the ratio. We use the ratio that contains the two sides involved: the opposite and the adjacent. Use SOH CAH TOA to pick it.
TOA stands for Tan, Opposite, Adjacent:
tan θ = OA (O = opposite, A = adjacent)
Put the numbers into the ratio. Replace θ with the angle and the sides with the lengths.
tan 27° = 14.2x
Multiply both sides by x. x is on the bottom of the fraction. Multiplying by x brings it to the top.
x × tan 27° = 14.2
Divide both sides by tan 27°. This leaves x on its own.
x = 14.2tan 27° = 14.2 ÷ tan 27°
Use the calculator (degree mode). First find tan 27°, then finish the calculation.
tan 27° = 0.50952…
x = 14.2 ÷ 0.50952… = 27.86906…
Round to 3 s.f.. Give the answer to the accuracy asked, with the unit.
x = 27.9 mm

Check. The other acute angle is 90° − 27° = 63°. From that angle the opposite and adjacent sides swap, so we use tan: x = 14.2 × tan 63° = 27.86906…, the same answer. ✓

Watch out. Check your calculator is in degree mode (DEG). The unknown is on the bottom, so you divide. A quick check: the hypotenuse must be the longest side.

Watch out

Label the sides from the angle you are using, not from the picture. The same side can be opposite for one angle and adjacent for the other. Also check that your calculator is in degree mode. If sin 30° does not give 0.5, it is not.

20.5

Trigonometry: finding angles

Edexcel 1MA1: G20
Grade 4-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Trigonometry: finding anglesComing soon

In the last section we knew an angle and wanted a side. Now we know two sides and want the angle. The method is almost the same, but at the end we use the inverse function.

Think about sin 30° = 0.5. Sin takes an angle (30°) and gives a ratio (0.5). The inverse function goes the other way: it takes the ratio and gives the angle. It is written sin⁻¹ and said “inverse sine” (or “arcsin”). So sin⁻¹(0.5) = 30°. On a calculator you press SHIFT (or 2ndF) and then sin, cos or tan. This is not the same as 1 ÷ sin. The little −1 is just a name.

We know O = 6.7 and H = 10, so use SOH: sin θ = 6.7 ÷ 10, then θ = sin⁻¹(0.67).

The method for finding an angle.

Label the two sides you know as O, A or H, measured from the angle you want.
Choose the ratio (SOH, CAH or TOA) that uses those two sides.
Write the ratio as a fraction with the numbers: for example tan θ = 5 ÷ 8.
Use the inverse button on the calculator: θ = tan⁻¹(5 ÷ 8).
Round sensibly and write the degree sign.

Do not round the fraction before pressing the inverse button, because the rounding error in the ratio makes the angle less accurate. Use the fraction key, or put the division in brackets.

Two checks help. First, an angle in a right-angled triangle is less than 90°. Second, the biggest side is opposite the biggest angle, so if the opposite side is long the angle should be large. We also use this method when the triangle is hidden in another shape, for example an isosceles triangle cut in half, or a rectangle with its diagonal drawn in.

Key ideas

To find an angle, use the inverse function: sin⁻¹, cos⁻¹ or tan⁻¹ (SHIFT then sin, cos or tan).
If sin θ = 0.5 then θ = sin⁻¹(0.5) = 30°.
Choose the ratio from the two sides you know.
Do not round the ratio before using the inverse.
Check: the angle must be less than 90° and the biggest side is opposite the biggest angle.

Worked example 20.5.1

Work out the size of angle x. Give your answer correct to 1 decimal place.

Method. Label the sides, pick SOH, CAH or TOA, write the ratio as a fraction, then use the inverse function to turn the ratio into an angle.

Label the sides. Stand at the angle we want. The hypotenuse is opposite the right angle. The opposite is across from the angle and the adjacent is next to it.
6.7 cm is the opposite.
10 cm is the hypotenuse.
Choose the ratio. We know the opposite and the hypotenuse, so we use the ratio that contains both. Use SOH CAH TOA.
SOH stands for Sin, Opposite, Hypotenuse:
sin θ = OH (O = opposite, H = hypotenuse)
Put the numbers into the ratio. Write the sides as a fraction (top = opposite, bottom = hypotenuse).
sin x = 6.710 = 0.67
Use the inverse function sin⁻¹. We know the value of sin x and want the angle. The inverse button (SHIFT or 2ndF, then sin) turns the ratio back into an angle.
x = sin⁻¹(6.710) = sin⁻¹(0.67) = 42.06706…°
Round to 1 d.p. Give the angle to the accuracy asked, with the degree sign.
x = 42.1°

Check. Work forwards: sin 42.1° = 0.670 and the ratio was 0.670. They agree. ✓

Watch out. To find an angle use the inverse button (sin⁻¹), not sin. Do not round the ratio before pressing the inverse button.

Worked example 20.5.2

Work out the size of angle x. Give your answer correct to 1 decimal place.

Method. Label the sides, pick SOH, CAH or TOA, write the ratio as a fraction, then use the inverse function to turn the ratio into an angle.

Label the sides. Stand at the angle we want. The hypotenuse is opposite the right angle. The opposite is across from the angle and the adjacent is next to it.
5 cm is the opposite.
8 cm is the adjacent.
Choose the ratio. We know the opposite and the adjacent, so we use the ratio that contains both. Use SOH CAH TOA.
TOA stands for Tan, Opposite, Adjacent:
tan θ = OA (O = opposite, A = adjacent)
Put the numbers into the ratio. Write the sides as a fraction (top = opposite, bottom = adjacent).
tan x = 58 = 0.625
Use the inverse function tan⁻¹. We know the value of tan x and want the angle. The inverse button (SHIFT or 2ndF, then tan) turns the ratio back into an angle.
x = tan⁻¹(58) = tan⁻¹(0.625) = 32.00538…°
Round to 1 d.p. Give the angle to the accuracy asked, with the degree sign.
x = 32.0°

Check. Work forwards: tan 32.0° = 0.625 and the ratio was 0.625. They agree. ✓

Watch out. To find an angle use the inverse button (tan⁻¹), not tan. Do not round the ratio before pressing the inverse button.

Worked example 20.5.3

Work out the size of angle x. Give your answer correct to 1 decimal place.

Method. Label the sides, pick SOH, CAH or TOA, write the ratio as a fraction, then use the inverse function to turn the ratio into an angle.

Label the sides. Stand at the angle we want. The hypotenuse is opposite the right angle. The opposite is across from the angle and the adjacent is next to it.
4.5 m is the adjacent.
9 m is the hypotenuse.
Choose the ratio. We know the adjacent and the hypotenuse, so we use the ratio that contains both. Use SOH CAH TOA.
CAH stands for Cos, Adjacent, Hypotenuse:
cos θ = AH (A = adjacent, H = hypotenuse)
Put the numbers into the ratio. Write the sides as a fraction (top = adjacent, bottom = hypotenuse).
cos x = 4.59 = 0.5
Use the inverse function cos⁻¹. We know the value of cos x and want the angle. The inverse button (SHIFT or 2ndF, then cos) turns the ratio back into an angle.
x = cos⁻¹(4.59) = cos⁻¹(0.5) = 60°
Round to 1 d.p. Give the angle to the accuracy asked, with the degree sign.
x = 60.0°

Check. Work forwards: cos 60.0° = 0.500 and the ratio was 0.500. They agree. ✓

Watch out. To find an angle use the inverse button (cos⁻¹), not cos. Do not round the ratio before pressing the inverse button.

Worked example 20.5.4

An isosceles triangle has two sides of length 10 cm and a base of 12 cm. Work out the size of the base angle x, correct to 1 decimal place.

Method. Label the sides, pick SOH, CAH or TOA, write the ratio as a fraction, then use the inverse function to turn the ratio into an angle.

Draw the line of symmetry. In an isosceles triangle the line of symmetry cuts the base in half at 90° and makes two right-angled triangles. We use one of them.
Half the base = 12 ÷ 2 = 6 cm. The slanting side is 10 cm (hypotenuse).
Label the sides. Stand at the angle we want. The hypotenuse is opposite the right angle. The opposite is across from the angle and the adjacent is next to it.
6 cm is the adjacent.
10 cm is the hypotenuse.
Choose the ratio. We know the adjacent and the hypotenuse, so we use the ratio that contains both. Use SOH CAH TOA.
CAH stands for Cos, Adjacent, Hypotenuse:
cos θ = AH (A = adjacent, H = hypotenuse)
Put the numbers into the ratio. Write the sides as a fraction (top = adjacent, bottom = hypotenuse).
cos x = 610 = 0.6
Use the inverse function cos⁻¹. We know the value of cos x and want the angle. The inverse button (SHIFT or 2ndF, then cos) turns the ratio back into an angle.
x = cos⁻¹(610) = cos⁻¹(0.6) = 53.13010…°
Round to 1 d.p. Give the angle to the accuracy asked, with the degree sign.
x = 53.1°

Check. Work forwards: cos 53.1° = 0.600 and the ratio was 0.600. They agree. ✓

Watch out. To find an angle use the inverse button (cos⁻¹), not cos. Do not round the ratio before pressing the inverse button.

Watch out

Use the inverse button, not the normal one. If you type sin(0.67) you get 0.0117…, not the angle. Also remember to put the degree sign in the answer. And if your calculator gives an error for sin⁻¹ or cos⁻¹, check that the number you typed is not bigger than 1: the hypotenuse must be on the bottom.

20.6

Angles of elevation and depression, and bearings

Edexcel 1MA1: G20, G15
Grade 4-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Angles of elevation and depression, and bearingsComing soon

Angle of elevation. When you look up at something, the angle between the horizontal line and your line of sight is the angle of elevation. Angle of depression. When you look down at something, the angle between the horizontal line and your line of sight is the angle of depression. Both angles are always measured from the horizontal, never from the vertical.

From O, the angle of elevation of the top T of the tower is θ. tan θ = 45 ÷ 60.

From the top of the cliff O, the angle of depression of the boat B is 25°. The angle at B is also 25° (alternate angles).

The horizontal line from the top of the cliff and the ground are parallel. So the angle of depression at the top and the angle of elevation of the top from the boat are alternate angles (Chapter 17), and they are equal. This is why we can use the angle at the bottom in the triangle, even though the question gives the angle at the top.

Bearings. A bearing is a direction measured clockwise from north, written with three digits. North is 000°, east is 090°, south is 180° and west is 270°. To go back, the back bearing is the original bearing plus or minus 180°. For example the bearing of B from A is 070°, so the bearing of A from B is 250°. Chapter 22 covers bearings in more detail. Here we use right-angled triangles to solve bearing problems.

Bearings are measured clockwise from north.

When a ship sails on a bearing, draw a north line at the start, mark the bearing angle clockwise from it, and then make a right-angled triangle with east-west and north-south lines. The north line is the side next to the bearing angle, and the east line is opposite it. So for a distance d on a bearing b:

distance east = d sin b distance north = d cos b

These two formulas work for every bearing. If the answer is negative, it means west (for east) or south (for north).

P to Q is 20 km on a bearing of 060°. East = 20 sin 60° = 17.3 km, north = 20 cos 60° = 10 km.

Key ideas

Elevation: looking up from the horizontal. Depression: looking down from the horizontal.
The angle of depression from the top equals the angle of elevation from the bottom (alternate angles).
Bearings: clockwise from north, three digits. Back bearing = bearing ± 180°.
Draw the right-angled triangle and use SOH CAH TOA or Pythagoras.

Worked example 20.6.1

From a point on level ground 60 m from the foot of a tower, the top of the tower is seen at an angle of elevation of θ. The tower is 45 m tall. Work out θ, correct to 1 decimal place.

Method. Label the sides, pick SOH, CAH or TOA, write the ratio as a fraction, then use the inverse function to turn the ratio into an angle.

Draw the right-angled triangle. The tower is vertical and the ground is horizontal, so they make a right angle. The angle of elevation is at the observer, between the ground and the line of sight.
Opposite the angle: tower height 45 m. Adjacent to the angle: 60 m along the ground.
Label the sides. Stand at the angle we want. The hypotenuse is opposite the right angle. The opposite is across from the angle and the adjacent is next to it.
45 m is the opposite.
60 m is the adjacent.
Choose the ratio. We know the opposite and the adjacent, so we use the ratio that contains both. Use SOH CAH TOA.
TOA stands for Tan, Opposite, Adjacent:
tan θ = OA (O = opposite, A = adjacent)
Put the numbers into the ratio. Write the sides as a fraction (top = opposite, bottom = adjacent).
tan x = 4560 = 0.75
Use the inverse function tan⁻¹. We know the value of tan x and want the angle. The inverse button (SHIFT or 2ndF, then tan) turns the ratio back into an angle.
x = tan⁻¹(4560) = tan⁻¹(0.75) = 36.86989…°
Round to 1 d.p. Give the angle to the accuracy asked, with the degree sign.
x = 36.9°

Check. Work forwards: tan 36.9° = 0.75 and the ratio was 0.75. They agree. ✓

Watch out. To find an angle use the inverse button (tan⁻¹), not tan. Do not round the ratio before pressing the inverse button.

Worked example 20.6.2

The angle of elevation of the top of a tree from a point 50 m away on level ground is 32°. Work out the height of the tree, correct to 1 decimal place.

Method. Label the sides, pick SOH, CAH or TOA, write the ratio with the numbers, then rearrange to get the unknown side.

Draw the right-angled triangle. The angle of elevation is 32° at the observer. The height of the tree is vertical, so it is opposite the angle. The ground distance is adjacent.
Adjacent = 50 m, opposite = h.
Label the sides. Stand at the angle θ. The hypotenuse is opposite the right angle. The opposite is across from θ and the adjacent is next to θ.
50 m is the adjacent.
h is the opposite.
Choose the ratio. We use the ratio that contains the two sides involved: the adjacent and the opposite. Use SOH CAH TOA to pick it.
TOA stands for Tan, Opposite, Adjacent:
tan θ = OA (O = opposite, A = adjacent)
Put the numbers into the ratio. Replace θ with the angle and the sides with the lengths.
tan 32° = h50
Multiply both sides by 50. h is being divided by 50. Multiplying both sides by 50 gets h on its own.
h = 50 × tan 32°
Use the calculator (degree mode). First find tan 32°, then finish the calculation.
tan 32° = 0.62486…
h = 50 × 0.62486… = 31.24346…
Round to 1 d.p.. Give the answer to the accuracy asked, with the unit.
h = 31.2 m

Check. The other acute angle is 90° − 32° = 58°. From that angle the opposite and adjacent sides swap, so we use tan: h = 50 ÷ tan 58° = 31.24346…, the same answer. ✓

Watch out. Check your calculator is in degree mode (DEG). The unknown is on the top, so you multiply. A quick check: the hypotenuse must be the longest side.

Worked example 20.6.3

From the top of a vertical cliff 80 m high, the angle of depression of a boat at sea is 25°. How far is the boat from the foot of the cliff? Give your answer correct to 1 decimal place.

Method. Label the sides, pick SOH, CAH or TOA, write the ratio with the numbers, then rearrange to get the unknown side.

Draw the right-angled triangle. The cliff is vertical, so the cliff and the sea level make a right angle. The angle of depression 25° at the top equals the angle at the boat (alternate angles).
At the boat the angle is 25°. The cliff height 80 m is opposite that angle. The distance x from the cliff is adjacent.
Label the sides. Stand at the angle θ. The hypotenuse is opposite the right angle. The opposite is across from θ and the adjacent is next to θ.
80 m is the opposite.
x is the adjacent.
Choose the ratio. We use the ratio that contains the two sides involved: the opposite and the adjacent. Use SOH CAH TOA to pick it.
TOA stands for Tan, Opposite, Adjacent:
tan θ = OA (O = opposite, A = adjacent)
Put the numbers into the ratio. Replace θ with the angle and the sides with the lengths.
tan 25° = 80x
Multiply both sides by x. x is on the bottom of the fraction. Multiplying by x brings it to the top.
x × tan 25° = 80
Divide both sides by tan 25°. This leaves x on its own.
x = 80tan 25° = 80 ÷ tan 25°
Use the calculator (degree mode). First find tan 25°, then finish the calculation.
tan 25° = 0.46630…
x = 80 ÷ 0.46630… = 171.56055…
Round to 1 d.p.. Give the answer to the accuracy asked, with the unit.
x = 171.6 m

Check. The other acute angle is 90° − 25° = 65°. From that angle the opposite and adjacent sides swap, so we use tan: x = 80 × tan 65° = 171.56055…, the same answer. ✓

Watch out. Check your calculator is in degree mode (DEG). The unknown is on the bottom, so you divide. A quick check: the hypotenuse must be the longest side.

Worked example 20.6.4

A ship sails 12 km due east from A to B, and then 5 km due south from B to C. Work out (a) the distance AC and (b) the bearing of A from C, to the nearest degree.

Method. The east and south journeys are at right angles, so use Pythagoras for the distance and tan for the angle. Then use the back bearing rule.

Draw the journey. Draw a north line at A. B is due east of A (bearing 090°), and C is due south of B (bearing 180° from B). The angle at B is 90°.
AB = 12 km, BC = 5 km.
Find the distance AC. AC is the hypotenuse of the right-angled triangle ABC.
AC² = 12² + 5² = 144 + 25 = 169, so AC = 13 km
Find angle BAC. At A: the opposite side is BC = 5 and the adjacent is AB = 12 (TOA).
tan BAC = 512 = 0.41666…
BAC = tan⁻¹(0.41666…) = 22.61986…°
Find the bearing of C from A. Bearings are measured clockwise from north. East is 090°, and C is further round (towards south) by angle BAC.
Bearing = 90° + 22.6° = 112.6°
Find the bearing of A from C. The back bearing is 180° more (or less).
112.6° + 180° = 292.6°, which is 293° to the nearest degree.

Check. Another way: at C the north line points towards B, so angle ACB = 90° − 22.6° = 67.4°. A is to the west of that line, so the bearing is 360° − 67.4° = 292.6°. ✓

Watch out. Draw a north line at each point you measure a bearing from. The bearing at C uses the north line at C.

Worked example 20.6.5

Q is 20 km from P on a bearing of 060°. How far east and how far north of P is Q? Give your answers correct to 1 decimal place.

Method. The path, the north line and the east line make a right-angled triangle. The bearing angle is between the north line and the path.

Draw the right-angled triangle. Draw a north line at P. The path to Q makes 60° with it. Go east and north to complete the right angle.
Hypotenuse = 20 km. The angle at P (between north and the path) = 60°.
Label from the 60° angle. The north side is next to the 60° angle (adjacent). The east side is across from it (opposite).
North = A, East = O, path = H
Find the distance east. H and O make SOH.
East = 20 × sin 60° = 20 × 0.86602… = 17.32050… km = 17.3 km
Find the distance north. H and A make CAH.
North = 20 × cos 60° = 20 × 0.5 = 10 km = 10.0 km

Check. Pythagoras: 17.3² + 10.0² = 399.3, which is 20² = 400. ✓

Watch out. The bearing is measured from north, so the north side is the adjacent, not the opposite.

Watch out

Bearings start from north and go clockwise, and need three figures (so 40° is written 040°). Draw a north line at every point where you measure a bearing. For the angle of depression, the angle is below the horizontal at the top, not the angle between the line of sight and the cliff.

20.7

Exact values of sin, cos and tan

Edexcel 1MA1: G21
Grade 5-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Exact values of sin, cos and tanComing soon

Most trigonometry answers are rounded decimals from a calculator. But for a few special angles, the ratios can be written exactly, using fractions and square roots. You need to know these exact values for 0°, 30°, 45°, 60° and 90°. They come up in non-calculator exam questions.

You do not have to learn them by heart. Two simple triangles give you all of them.

The 45° triangle. Take a square of side 1 and cut it along a diagonal. You get a right-angled triangle with two equal sides of length 1 and two angles of 45°. By Pythagoras the hypotenuse is √(1² + 1²) = √2.

A square of side 1 cut in half: a right-angled triangle with sides 1, 1 and √2 and two 45° angles.

The 30° and 60° triangle. Take an equilateral triangle of side 2 and cut it in half down the middle. You get a right-angled triangle with angles 30°, 60° and 90°. The shortest side is 1 (half of the base 2), the hypotenuse is 2, and the third side is √(2² − 1²) = √3.

An equilateral triangle of side 2 cut in half gives a triangle with sides 1, √3 and 2, with angles 30°, 60°, 90°.

Now use SOH CAH TOA. For example, in the 30°–60°–90° triangle, for the 30° angle the opposite side is 1 and the hypotenuse is 2, so sin 30° = 1÷2. For the 60° angle the opposite side is √3, so sin 60° = √3÷2. For tan 30° we get 1÷√3. We usually remove the square root from the bottom by multiplying the top and bottom by √3, which gives √3÷3. This is called rationalising the denominator.

The values for 0° and 90° come from a triangle that has been squashed flat (0°) or stretched upright (90°). Here is the complete table:

θ	0°	30°	45°	60°	90°
sin θ	0	12	√22	√32	1
cos θ	1	√32	√22	12	0
tan θ	0	√33	1	√3	undefined

A pattern to help you remember sin: the values are √0÷2, √1÷2, √2÷2, √3÷2, √4÷2, which simplify to 0, ½, √2÷2, √3÷2, 1. The cos row is the sin row backwards. And tan = sin ÷ cos.

Key ideas

sin 30° = ½ cos 60° = ½ sin 60° = cos 30° = √3÷2 sin 45° = cos 45° = √2÷2.
tan 30° = √3÷3 (or 1÷√3), tan 45° = 1, tan 60° = √3. tan 90° is undefined.
sin 0° = 0, cos 0° = 1, tan 0° = 0; sin 90° = 1, cos 90° = 0.
Rationalise: 1÷√2 = √2÷2 and 1÷√3 = √3÷3.
Exact answers keep the √ and the fractions. Do not use decimals.

Worked example 20.7.1

Write down the exact value of sin 60°.

Method. Draw the special triangle for the angle, label the sides with exact lengths, then write the ratio.

Use the special triangle. For 60° we use half of an equilateral triangle of side 2 (sides 1, √3 and 2). These triangles give exact side lengths.
Opposite = √3, adjacent = 1, hypotenuse = 2 (for the 60° angle).
Write the ratio for sin. SOH: sin = opposite ÷ hypotenuse.
sin 60° = √32
Simplify. Check whether the fraction can be written more simply.
sin 60° = √32

Check. Calculator: sin 60° = 0.86602… and the exact value is 0.86602…. They match. ✓

Watch out. You can learn the table, but if you forget a value you can always rebuild it from the two triangles.

Worked example 20.7.2

Write down the exact value of tan 30°. Give your answer with a rational denominator.

Method. Draw the special triangle for the angle, label the sides with exact lengths, then write the ratio.

Use the special triangle. For 30° we use half of an equilateral triangle of side 2 (sides 1, √3 and 2). These triangles give exact side lengths.
Opposite = 1, adjacent = √3, hypotenuse = 2 (for the 30° angle).
Write the ratio for tan. TOA: tan = opposite ÷ adjacent.
tan 30° = 1√3
Rationalise the denominator. We do not leave a square root on the bottom. Multiply the top and the bottom by √3.
1√3 = 1 × √3√3 × √3 = √33

Check. Calculator: tan 30° = 0.57735… and the exact value is 0.57735…. They match. ✓

Watch out. You can learn the table, but if you forget a value you can always rebuild it from the two triangles.

Worked example 20.7.3

Work out the exact value of sin 60° × cos 30°.

Method. Replace each trigonometric ratio by its exact value, then work out the answer using fractions and surds.

Write down each exact value. Use the table or the special triangles. Do not use decimals.
sin 60° = √32
cos 30° = √32
Put them into the calculation. Replace each trig value by its exact value.
sin 60° × cos 30° = √32 × √32
Work it out exactly. To multiply fractions, multiply the tops and multiply the bottoms. Then simplify.
√32 × √32 = 34

Check. Decimal check: 0.86602… × 0.86602… = 0.75, and 34 = 0.75. ✓

Watch out. Keep surds and fractions. Writing 0.866 for √3 ÷ 2 gives an approximate answer, which does not score full marks in an exact question.

Worked example 20.7.4

The hypotenuse of this right-angled triangle is 10 cm. Work out the exact length x.

Key words. An exact value is written with fractions and square roots (surds) instead of a rounded decimal. To rationalise a fraction means to remove the square root from the bottom by multiplying the top and bottom by that square root. In a right-angled triangle the hypotenuse is the longest side (opposite the right angle). Measured from the angle θ: the opposite side is directly across from θ, and the adjacent side is next to θ (it is not the hypotenuse). A trigonometric ratio compares two of the sides.

Method. Use SOH CAH TOA as usual, but put in the exact value of sin, cos or tan for the special angle so the answer stays exact.

Label the sides and choose the ratio. We have the hypotenuse and want the opposite.
SOH stands for Sin, Opposite, Hypotenuse:
sin θ = OH (O = opposite, H = hypotenuse)
Put the numbers in. The angle is one of the special angles, so use its exact value.
sin 60° = x10
sin 60° = √32
Multiply both sides by 10. x is divided by 10.
x = 10 × √32 = 5√3

Check. Decimal check: 10 × sin 60° = 8.66025…, and 5√3 = 8.66025…. ✓

Watch out. If the unknown is on the bottom, dividing by a surd is awkward. Flip the exact value instead and multiply.

Worked example 20.7.5

Work out the size of the acute angle x when cos x = √2 ÷ 2.

Method. Look for the exact value √22 in the cos row of the table of exact values. The angle in that column is the answer.

Write the equation. We know the exact value of cos x.
cos x = √22
Look along the cos row. Check each angle until the value matches. The values in the row are 1, √32, √22, 12, 0.
cos 45° = √22
Write the angle. An acute angle in a right-angled triangle is between 0° and 90°.
x = 45°

Check. Calculator: cos⁻¹(0.70710…) = 45°. ✓

Watch out. This is an exact answer, so give it as a whole number of degrees, not a rounded decimal.

Watch out

Do not confuse the two special values: sin 30° = ½ but sin 60° = √3÷2 (about 0.87). A quick check: sin gets bigger as the angle gets bigger, so sin 60° must be bigger than sin 30°. Also, tan 45° = 1 exactly, and tan 90° has no value.

20.8

The sine rule

Edexcel 1MA1: G22
Grade 6-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: The sine ruleComing soon

SOH CAH TOA only works in right-angled triangles. The sine rule and the cosine rule work in any triangle. In triangle ABC we use a capital letter for each angle and the matching small letter for the side opposite it: side a is opposite angle A, side b is opposite angle B and side c is opposite angle C.

Each small letter is the side opposite the capital letter of the same name.

The sine rule.

asin A = bsin B = csin C

It says: each side divided by the sine of the angle opposite it gives the same number. You only need two of the three fractions at a time. To find an angle it is easier to turn the rule upside down: sin A ÷ a = sin B ÷ b = sin C ÷ c.

Why it works. Draw the perpendicular height h from C to the side AB. In the right-angled triangle on the left, sin A = h ÷ b, so h = b sin A. In the right-angled triangle on the right, sin B = h ÷ a, so h = a sin B. Both equal h, so b sin A = a sin B. Dividing both sides by sin A sin B gives a ÷ sin A = b ÷ sin B.

The height h = b sin A and also h = a sin B.

When to use the sine rule. You need a matching pair: a side and the angle opposite it, that you know completely. Then:

know two angles and any side (AAS or ASA): find another side. If only two angles are given, find the third angle first (the three angles add to 180°).
know two sides and an angle that is not between them (SSA, where the angle is opposite one of the sides): find another angle.

If you know two sides and the angle between them, or all three sides, there is no matching pair, so you use the cosine rule instead (next section).

The ambiguous case. When you use the sine rule to find an angle, the inverse sine only gives an acute angle. But sin x = sin (180° − x), so the obtuse angle 180° − x has the same sine. Sometimes both angles make a real triangle. For example, with angle A = 30°, side b = 10 cm and side a = 6 cm, the side BC can swing to meet the base line at two different places, B₁ and B₂. You get two different triangles from the same information.

Same angle (30°) and same sides (10 cm and 6 cm), but two possible triangles: AB₁C and AB₂C.

To decide: work out the obtuse possibility 180° − x. If the known angle plus this obtuse angle is less than 180°, a second triangle exists. This only happens when the side opposite the known angle is shorter than the other known side. If the question says the angle is acute (or obtuse), or the diagram shows it, choose that one.

Key ideas

Sine rule: a ÷ sin A = b ÷ sin B = c ÷ sin C. For angles: sin A ÷ a = sin B ÷ b = sin C ÷ c.
Use it when you have a matching pair (side and opposite angle).
Two angles given? Find the third angle first (angles add to 180°).
Finding an angle: the other possible angle is 180° − (calculator answer). Check it fits in the triangle.

Worked example 20.8.1

Work out the length x. Give your answer correct to 1 decimal place.

Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.

Method. The sine rule: each side divided by the sine of its opposite angle gives the same number. Use one complete pair to find the missing side.

Match each side with the angle opposite it. The sine rule links a side with the angle opposite that side. We need one matching pair that we know completely, and one pair with the unknown.
Known pair: side a = 8.5 cm is opposite angle A = 40°.
Wanted pair: side b is opposite angle B = 65°.
Write the sine rule with the unknown on top. Putting the side we want on top means we only need to multiply at the end.
bsin B = asin A
bsin 65° = 8.5sin 40°
Multiply both sides by sin 65°. This gets b on its own.
b = 8.5 × sin 65°sin 40°
Use the calculator (degree mode). Find the two sines, then multiply and divide.
sin 65° = 0.90630… and sin 40° = 0.64278…
b = 8.5 × 0.90630… ÷ 0.64278… = 11.98469…
Round to 1 d.p.. Give the answer to the accuracy asked, with the unit.
b = 12.0 cm

Check. Each side ÷ sine of its opposite angle should be the same number: 8.5 ÷ sin 40° = 13.2 and 12.0 ÷ sin 65° = 13.2. ✓

Watch out. Always pair a side with the angle opposite it (not next to it). Draw arrows from each angle to the side across from it if you are unsure.

Worked example 20.8.2

Work out the length x. Give your answer correct to 1 decimal place.

Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.

Method. The sine rule: each side divided by the sine of its opposite angle gives the same number. Use one complete pair to find the missing side.

Find the third angle. The angles in a triangle add up to 180°. We need angle C, because it is opposite the side we want.
C = 180° − 48° − 71° = 61°
Match each side with the angle opposite it. The sine rule links a side with the angle opposite that side. We need one matching pair that we know completely, and one pair with the unknown.
Known pair: side a = 9.2 cm is opposite angle A = 48°.
Wanted pair: side c is opposite angle C = 61°.
Write the sine rule with the unknown on top. Putting the side we want on top means we only need to multiply at the end.
csin C = asin A
csin 61° = 9.2sin 48°
Multiply both sides by sin 61°. This gets c on its own.
c = 9.2 × sin 61°sin 48°
Use the calculator (degree mode). Find the two sines, then multiply and divide.
sin 61° = 0.87461… and sin 48° = 0.74314…
c = 9.2 × 0.87461… ÷ 0.74314… = 10.82763…
Round to 1 d.p.. Give the answer to the accuracy asked, with the unit.
c = 10.8 cm

Check. Each side ÷ sine of its opposite angle should be the same number: 9.2 ÷ sin 48° = 12 and 10.8 ÷ sin 61° = 12. ✓

Watch out. Always pair a side with the angle opposite it (not next to it). Draw arrows from each angle to the side across from it if you are unsure.

Worked example 20.8.3

In this triangle angle x is acute. Work out the size of angle x, correct to 1 decimal place.

Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.

Method. Use the sine rule with the sines on top: sin(wanted angle) over its opposite side equals sin(known angle) over its opposite side. Then use the inverse sine.

Match each side with the angle opposite it. The sine rule links a side with the angle opposite it. We have one complete pair, and the side opposite the angle we want.
Known pair: side a = 10 cm is opposite angle A = 52°.
Wanted: angle B, which is opposite side b = 7 cm.
Write the sine rule with the sines on top. When we want an angle it is easier to put the sines on top.
sin Bb = sin Aa
sin B7 = sin 52°10
Multiply both sides by 7. This gets sin B on its own.
sin B = 7 × sin 52°10 = 0.55160…
Use the inverse sine to find B. The inverse button turns a sine value back into an angle.
B = sin⁻¹(0.55160…) = 33.47736…°
Look for a second (obtuse) answer. Sine gives the same value for an acute angle and for 180° minus it. The obtuse angle only works if the angles still fit inside 180°.
Second angle = 180° − 33.47736…° = 146.52263…°
Angles so far: 52° + 146.52263…° = 198.52263…°, which is more than 180°, so it is impossible. Only the acute answer works.
Round to 1 d.p. Only one answer is possible.
B = 33.5°

Check. Each sine ÷ its opposite side should match: sin 33.5° ÷ 7 = 0.079 and sin 52° ÷ 10 = 0.079. ✓

Watch out. The inverse sine always gives an acute angle. If the triangle might have an obtuse angle, remember the second answer 180° minus the first.

Worked example 20.8.4

In triangle ABC, angle A = 30°, AC = 10 cm and BC = 6 cm. Work out the possible sizes of angle B, correct to 1 decimal place.

Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.

Method. Use the sine rule with the sines on top: sin(wanted angle) over its opposite side equals sin(known angle) over its opposite side. Then use the inverse sine.

Match each side with the angle opposite it. The sine rule links a side with the angle opposite it. We have one complete pair, and the side opposite the angle we want.
Known pair: side a = 6 cm is opposite angle A = 30°.
Wanted: angle B, which is opposite side b = 10 cm.
Write the sine rule with the sines on top. When we want an angle it is easier to put the sines on top.
sin Bb = sin Aa
sin B10 = sin 30°6
Multiply both sides by 10. This gets sin B on its own.
sin B = 10 × sin 30°6 = 0.83333…
Use the inverse sine to find B. The inverse button turns a sine value back into an angle.
B = sin⁻¹(0.83333…) = 56.44269…°
Look for a second (obtuse) answer. Sine gives the same value for an acute angle and for 180° minus it. A second triangle is possible if the angles still fit inside 180°.
Second angle = 180° − 56.44269…° = 123.55730…°
Angles so far: 30° + 123.55730…° = 153.55730…°, which is less than 180°, so this second triangle can exist.
Round to 1 d.p. and state both answers. Both triangles fit the information given (the ambiguous case).
B = 56.4° or B = 123.6°

Check. Each sine ÷ its opposite side should match: sin 56.4° ÷ 10 = 0.083 and sin 30° ÷ 6 = 0.083. ✓

Watch out. The inverse sine always gives an acute angle. If the triangle might have an obtuse angle, remember the second answer 180° minus the first.

Watch out

Match each side with the angle opposite it. A very common mistake is to use the angle at one end of the side. And when you find an angle, always ask: could the obtuse angle also work? Do not round the sine values in the middle of the calculation: keep them on the calculator.

20.9

The cosine rule

Edexcel 1MA1: G22
Grade 6-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: The cosine ruleComing soon

The sine rule needs a matching pair (a side and its opposite angle). What if you do not have one? For example, if you know two sides and the angle between them, or all three sides. Then you use the cosine rule.

a² = b² + c² − 2bc cos A

Here a is the side opposite angle A, and b and c are the two other sides. The angle A is the angle between sides b and c. It looks like Pythagoras’ theorem with a correction term. If A = 90° then cos A = 0, the last term disappears and we get a² = b² + c². If A is bigger than 90°, cos A is negative, so −2bc cos A is positive and a is longer than Pythagoras would give. That makes sense: a wider angle gives a longer opposite side.

The cosine rule: the side a is opposite angle A, and A is between sides b and c.

Finding a side (two sides and the angle between them, SAS). Substitute the numbers, work out 2bc cos A first as one number, subtract it from b² + c², and take the square root.

Finding an angle (three sides, SSS). Rearrange the rule to make cos A the subject:

cos A = b² + c² − a²2bc

Then use the inverse cosine. If the top is negative, the angle is obtuse, and the calculator gives it correctly (unlike the sine rule, there is no second answer).

Which rule? Look at what is given. A matching side and angle pair, plus one more value: sine rule. Two sides and the angle between them (SAS), or three sides (SSS): cosine rule. Start with the cosine rule when you cannot find a matching pair.

Key ideas

Cosine rule: a² = b² + c² − 2bc cos A.
To find an angle: cos A = (b² + c² − a²) ÷ 2bc.
Use it for SAS (two sides and the angle between) and SSS (three sides).
Work out 2bc cos A before subtracting.
cos of an obtuse angle is negative.

Worked example 20.9.1

Work out the length x. Give your answer correct to 1 decimal place.

Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.

Method. The cosine rule finds the third side of a triangle when two sides and the angle between them are known: a² = b² + c² − 2bc cos A.

Spot the angle and the two sides around it. The cosine rule is used when we know two sides and the angle between them. The side we want is the one opposite that angle.
Known: b = 7 cm, c = 9 cm, angle A = 50° between them.
Wanted: a, opposite angle A.
Write the cosine rule and put the numbers in. Square the wanted side on the left; square the other two sides on the right, then take away 2 × side × side × cos(angle).
a² = b² + c² − 2bc cos A
a² = 7² + 9² − 2 × 7 × 9 × cos 50°
Work out the two squares and add. This is the first part, b² + c².
7² + 9² = 49 + 81 = 130
Work out the last part on its own. Multiply everything together first. Do this before subtracting (BIDMAS: multiplication comes before subtraction).
cos 50° = 0.64278…
2 × 7 × 9 × 0.64278… = 80.99123…
Subtract. This gives a².
a² = 130 − 80.99123… = 49.00876…
Square root, then round to 1 d.p.. We want a, not a².
a = √49.00876… = 7.00062…
a = 7.0 cm

Check. One side of a triangle is between the difference and the sum of the other two: 2 is less than 7.0, which is less than 16. Angle A is less than 90°, so a² should be less than b² + c²: 49.00876… is less than 130. ✓

Watch out. Do 2bc cos A first and then subtract it. A very common mistake is to work out (b² + c² − 2bc) first. The rule has a minus sign, even for an obtuse angle (cos of an obtuse angle is negative).

Worked example 20.9.2

Work out the length x. Give your answer correct to 1 decimal place.

Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.

Method. The cosine rule finds the third side of a triangle when two sides and the angle between them are known: a² = b² + c² − 2bc cos A.

Spot the angle and the two sides around it. The cosine rule is used when we know two sides and the angle between them. The side we want is the one opposite that angle.
Known: b = 6 cm, c = 8 cm, angle A = 120° between them.
Wanted: a, opposite angle A.
Write the cosine rule and put the numbers in. Square the wanted side on the left; square the other two sides on the right, then take away 2 × side × side × cos(angle).
a² = b² + c² − 2bc cos A
a² = 6² + 8² − 2 × 6 × 8 × cos 120°
Work out the two squares and add. This is the first part, b² + c².
6² + 8² = 36 + 64 = 100
Work out the last part on its own. Multiply everything together first. Do this before subtracting (BIDMAS: multiplication comes before subtraction).
cos 120° = −0.5
2 × 6 × 8 × −0.5 = −48
Subtract. This gives a².
a² = 100 − −48 = 148
Square root, then round to 1 d.p.. We want a, not a².
a = √148 = 12.16552…
a = 12.2 cm

Check. One side of a triangle is between the difference and the sum of the other two: 2 is less than 12.2, which is less than 14. Angle A is more than 90°, so a² should be more than b² + c²: 148 is more than 100. ✓

Worked example 20.9.3

Work out the size of angle x. Give your answer correct to 1 decimal place.

Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.

Method. When all three sides are known, rearrange the cosine rule to cos A = (b² + c² − a²) ÷ 2bc, then use the inverse cosine.

Spot the angle we want and the side opposite it. With three sides known we can find any angle. Call the side opposite the wanted angle a.
Wanted: angle A, opposite the side a = 11 cm.
The other two sides are b = 7 cm and c = 9 cm.
Write the cosine rule and rearrange it for cos A. We want cos A on its own, so add 2bc cos A to both sides, subtract a² from both sides, then divide by 2bc.
a² = b² + c² − 2bc cos A
2bc cos A = b² + c² − a²
cos A = b² + c² − a²2bc
Put the numbers in the top and the bottom. Work the top and the bottom out separately so you do not make a slip.
Top: 7² + 9² − 11² = 49 + 81 − 121 = 9
Bottom: 2 × 7 × 9 = 126
Divide to find cos A. This gives the cosine of the angle as a decimal.
cos A = 9126 = 0.07142…
Use the inverse cosine. The inverse button (cos⁻¹) turns the cosine value back into an angle. (A negative cosine gives an obtuse angle, which is correct.)
A = cos⁻¹(0.07142…) = 85.90395…°
Round to 1 d.p. Give the angle to the accuracy asked.
A = 85.9°

Check. Put the angle back in the cosine rule: 7² + 9² − 2 × 7 × 9 × cos 85.9° = 121.0, and 11² = 121.0. ✓

Watch out. In the top line the side a opposite the wanted angle is the one that is subtracted. Work out the top and the bottom separately, and do not divide until both are finished.

Worked example 20.9.4

Work out the size of the largest angle x in this triangle. Give your answer correct to 1 decimal place.

Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.

Method. When all three sides are known, rearrange the cosine rule to cos A = (b² + c² − a²) ÷ 2bc, then use the inverse cosine.

Spot the angle we want and the side opposite it. With three sides known we can find any angle. Call the side opposite the wanted angle a.
Wanted: angle A, opposite the side a = 10 cm.
The other two sides are b = 5 cm and c = 6 cm.
Write the cosine rule and rearrange it for cos A. We want cos A on its own, so add 2bc cos A to both sides, subtract a² from both sides, then divide by 2bc.
a² = b² + c² − 2bc cos A
2bc cos A = b² + c² − a²
cos A = b² + c² − a²2bc
Put the numbers in the top and the bottom. Work the top and the bottom out separately so you do not make a slip.
Top: 5² + 6² − 10² = 25 + 36 − 100 = −39
Bottom: 2 × 5 × 6 = 60
Divide to find cos A. This gives the cosine of the angle as a decimal.
cos A = −3960 = −0.65
Use the inverse cosine. The inverse button (cos⁻¹) turns the cosine value back into an angle. (A negative cosine gives an obtuse angle, which is correct.)
A = cos⁻¹(−0.65) = 130.54160…°
Round to 1 d.p. Give the angle to the accuracy asked.
A = 130.5°

Check. Put the angle back in the cosine rule: 5² + 6² − 2 × 5 × 6 × cos 130.5° = 100.0, and 10² = 100.0. ✓

Watch out. In the top line the side a opposite the wanted angle is the one that is subtracted. Work out the top and the bottom separately, and do not divide until both are finished.

Watch out

Do the multiplication 2bc cos A first. Do not work out (b² + c² − 2bc) first and then multiply by cos A. And the angle must be the one between the two sides that are squared and added. If you find an obtuse angle, cos A is negative, so be careful with the signs.

20.10

Area of a triangle: ½ab sin C

Edexcel 1MA1: G23
Grade 6-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Area of a triangle: ½ab sin CComing soon

You already know that the area of a triangle is ½ × base × height. But sometimes the height is not given. If you know two sides and the angle between them, you can still find the area.

Area = ½ ab sin C

Here a and b are two sides and C is the angle between them.

Why it works. Take a as the base. Draw the height h from the opposite vertex. In the right-angled triangle, the hypotenuse is b and the angle at C is C, so sin C = h ÷ b, which gives h = b sin C. Then area = ½ × base × height = ½ × a × b sin C. That is the formula.

The height is h = b sin C, so the area is ½ × a × b sin C.

The formula can be used to find the area, but it can also be rearranged to find a side or an angle when the area is known. To find an angle, you get sin C. As for the sine rule, there are two angles between 0° and 180° with the same sine, an acute one and an obtuse one (180° minus the acute one). Read the question to see which is wanted, or give both.

The formula is also useful for polygons. A regular polygon with n sides splits into n identical isosceles triangles meeting at the centre. Each has two equal sides (the radius) and an angle of 360° ÷ n between them, so its area is ½ r² sin(360° ÷ n).

A regular octagon with radius 10 cm: 8 triangles, each with two sides of 10 cm and an angle of 45° between them.

Key ideas

Area = ½ ab sin C, where C is the angle between sides a and b.
To find an angle: sin C = 2 × area ÷ (ab). Remember the possible obtuse angle 180° − C.
To find a side, rearrange: a = 2 × area ÷ (b sin C).
Regular polygons: n triangles, each with area ½ r² sin(360° ÷ n).

Worked example 20.10.1

Work out the area of the triangle. Give your answer correct to 1 decimal place.

Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.

Method. Area of any triangle = ½ × a × b × sin C, where C is the angle between sides a and b.

Spot the two sides and the angle between them. The formula needs two sides and the included angle (the angle between those two sides).
Sides: 8 cm and 11 cm. Included angle C = 40°.
Write the formula and put the numbers in. Area = half × one side × the other side × sine of the angle between them.
Area = ½ ab sin C
Area = ½ × 8 × 11 × sin 40°
Use the calculator. Find the sine first, then multiply everything together.
sin 40° = 0.64278…
Area = 0.5 × 8 × 11 × 0.64278… = 28.28265…
Round to 1 d.p. and write the units. Area is measured in square units.
Area = 28.3 cm²

Check. Another way: take 8 as the base. The height is 11 × sin 40° = 7.07066…. Area = ½ × 8 × 7.07066… = 28.28265…. ✓

Watch out. The angle must be the one between the two sides you use. If it is not between them, find the angle you need first (angles in a triangle add up to 180°).

Worked example 20.10.2

A triangle has two sides of length 8 cm and 11 cm, and an area of 30 cm². The angle between the two sides is C. Work out the two possible sizes of C, correct to 1 decimal place.

Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.

Method. Put the area into Area = ½ ab sin C, rearrange to find sin C, then use the inverse sine.

Write the area formula and put the numbers in. The area is ½ × a × b × sin C, and here the angle C is unknown.
30 = ½ × 8 × 11 × sin C
Work out ½ × a × b. This is a plain number, so we can combine it.
½ × 8 × 11 = 44
30 = 44 sin C
Divide both sides by 44. This gets sin C on its own.
sin C = 3044 = 0.68181…
Use the inverse sine. The inverse button turns the sine back into an angle.
C = sin⁻¹(0.68181…) = 42.98588…°
Think about a second (obtuse) angle. Sine of an obtuse angle equals sine of 180° minus it, so both give the same area.
Second angle = 180° − 42.98588…° = 137.01411…°
Round to 1 d.p. Both angles give the same area, so both are answers.
C = 43.0° or 137.0°

Check. Work forwards: ½ × 8 × 11 × sin 43.0° = 30.0 cm², close to 30. ✓

Watch out. The inverse sine only gives the acute angle. An obtuse angle gives the same sine, so check whether the question wants it.

Worked example 20.10.3

A triangle has an area of 35 cm². Two of its sides are a and 10 cm, and the angle between them is 48°. Work out the length of side a, correct to 1 decimal place.

Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.

Method. Put the area into Area = ½ ab sin C and rearrange to find the missing side.

Write the area formula and put the numbers in. The area is ½ × a × b × sin C, and here one side is unknown.
35 = ½ × a × 10 × sin 48°
Work out the known numbers. Combine ½, the side and sin C into one number.
sin 48° = 0.74314…
½ × 10 × 0.74314… = 3.71572…
35 = 3.71572… a
Divide both sides by 3.71572…. This leaves a on its own.
a = 35 ÷ 3.71572… = 9.41942…
Round to 1 d.p. Give the unit.
a = 9.4 cm

Check. Work forwards: ½ × 9.4 × 10 × sin 48° = 34.9 cm², close to 35. ✓

Watch out. Be careful rearranging: divide by everything that multiplies the unknown side (the ½, the other side and sin C).

Worked example 20.10.4

A regular octagon has a distance of 10 cm from its centre to each corner. Work out its area, correct to 1 decimal place.

Key words. In triangle ABC the side a is opposite angle A, side b is opposite angle B and side c is opposite angle C. The included angle is the angle between two given sides.

Method. A regular octagon is made of 8 identical triangles. Find the area of one triangle with ½ ab sin C and multiply by 8.

Find the angle at the centre. The 8 angles at the centre make a full turn of 360°.
360° ÷ 8 = 45°
Find the area of one triangle. Two sides are the radius 10 cm, and the angle between them is 45°.
½ × 10 × 10 × sin 45° = 50 × 0.70710… = 35.35533… cm²
Multiply by 8. There are 8 identical triangles.
8 × 35.35533… = 282.84271…
Round.
Area = 282.8 cm²

Check. Exact version: 8 × 50 × √2÷2 = 200√2 = 282.84271…. ✓

Watch out. Multiply by the number of triangles at the end.

Watch out

The angle in the formula must be the one between the two sides you use. If it is not, you must find the angle you need first, using the fact that angles in a triangle add to 180°. Also remember the units: area is in square units, such as cm².

20.11

Trigonometry and Pythagoras in three dimensions, and combined problems

Edexcel 1MA1: G20, G22, G23
Grade 7-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Trigonometry and Pythagoras in three dimensions, and combined problemsComing soon

Many solids contain right-angled triangles. The trick for every 3D problem is the same: find a right-angled triangle inside the solid, draw it on its own as a flat picture, and use Pythagoras or SOH CAH TOA. Often you need to use two triangles one after the other.

The space diagonal of a cuboid. In the cuboid below, the line AG goes from one bottom corner to the opposite top corner. First find AC, the diagonal of the base, using triangle ABC (right angle at B). Then use triangle ACG (right angle at C, because CG is vertical) to find AG.

Triangle ACG is right-angled at C. AC² = 8² + 6², then AG² = AC² + 5².

Putting the two steps together gives the formula AG² = l² + w² + h², where l, w and h are the length, width and height. You must still understand how it arises, and show the two-triangle method in an exam if the question asks you to “show that”.

The angle between a line and a plane. The angle between a line and a flat surface (a plane) is the angle between the line and its shadow (its projection) on the plane. It is the smallest angle that the line makes with the plane. To find the angle between AG and the base ABCD: the shadow of AG on the base is AC (look straight down on AG). So the angle is the angle GAC, in the right-angled triangle ACG.

The angle between AG and the base is the angle θ between AG and its shadow AC.

Pyramids. A pyramid with a rectangular base has its apex V directly above the centre M of the base (this is the usual case). The centre M is halfway along both diagonals. The right-angled triangle V M A (right angle at M) contains the height VM, the half-diagonal AM and the sloping edge VA.

Triangle VMA is right-angled at M. The angle θ between the edge VA and the base is the angle VAM.

Combined problems. Some problems need several tools. A good plan: (1) draw each triangle separately and mark what you know; (2) decide, for each triangle, whether it is right-angled (Pythagoras, SOH CAH TOA) or not (sine rule, cosine rule, area formula); (3) work out an unknown that lets you move to the next triangle; (4) keep full calculator values until the end. Bearings, angles of elevation and 3D shapes all give you triangles of this kind.

Key ideas

3D: find a right-angled triangle, draw it flat, and label it with the lengths you know.
Space diagonal of a cuboid: AG² = l² + w² + h² (found in two steps).
The angle between a line and a plane = the angle between the line and its shadow on the plane.
Pyramid: the apex is above the centre M of the base; use the triangle with the height VM.
Not right-angled? Use the sine rule, the cosine rule or ½ab sin C.

Worked example 20.11.1

ABCDEFGH is a cuboid with AB = 8 cm, BC = 6 cm and CG = 5 cm. Work out the length of the diagonal AG, correct to 1 decimal place.

Method. Use Pythagoras’ theorem twice: first across the base to find the base diagonal, then upwards with the height to find the 3D diagonal.

Find the diagonal of the base first. AC lies along the base. Triangle ABC is right-angled at B, with AB = 8 cm and BC = 6 cm, so we can use Pythagoras.
AC² = 8² + 6² = 64 + 36 = 100
Now use the vertical edge. CG stands straight up from C, so triangle ACG is right-angled at C. The 3D diagonal AG is its hypotenuse. We can use the exact value AC² = 100 without finding AC first.
AG² = AC² + CG² = 100 + 5² = 100 + 25 = 125
Take the square root and round to 1 d.p. We want AG, not AG².
AG = √125 = 11.18033…
AG = 11.2 cm

Check. The shortcut: AG² = 8² + 6² + 5² = 64 + 36 + 25 = 125. ✓

Watch out. Do not add the three edges. Add their squares, and then take one square root at the end. Always find a right angle to work in, and draw the triangle separately.

Worked example 20.11.2

For the same cuboid, work out the angle between the diagonal AG and the base ABCD, correct to 1 decimal place.

Method. The angle between a line and a plane is the angle between the line and its shadow (projection) on the plane. Find the shadow, then use trigonometry in the right-angled triangle.

Find the projection of the line onto the base. The angle between a line and a plane is the angle between the line and its shadow on the plane. The shadow of AG on the base is AC. The angle we want is angle GAC.
Triangle ACG is right-angled at C.
Find the length of the shadow, AC. Use Pythagoras in the base triangle.
AC² = 8² + 6² = 100
AC = 10 cm
Label triangle ACG for trigonometry. From angle GAC: the opposite side is CG = 5 cm and the adjacent side is AC = 10 cm. Opposite and adjacent means TOA.
tan GAC = oppositeadjacent = 510
Use the inverse tangent. The inverse button turns the tangent value into an angle.
tan GAC = 0.5
angle GAC = tan⁻¹(0.5) = 26.56505…°
Round to 1 d.p.
Angle = 26.6°

Check. Use sine instead: sin GAC = 5 ÷ AG, where AG = √(64 + 36 + 25) = 11.18033…. 5 ÷ 11.18033… = 0.44721…, and sin 26.6° = 0.448. ✓

Watch out. The angle with the base is measured to the base diagonal AC, not to an edge of the base like AB.

Worked example 20.11.3

VABCD is a pyramid. The base ABCD is a square of side 10 cm and the apex V is 12 cm vertically above the centre M of the base. Work out the length of the edge VA, correct to 1 decimal place.

Method. Draw the right-angled triangle inside the pyramid that contains the height: apex, centre of the base and a corner. Use Pythagoras’ theorem.

Find the diagonal of the base. The apex is directly above the centre M of the base. M is the middle of the base diagonal AC. Triangle ABC is right-angled at B.
AC² = 10² + 10² = 100 + 100 = 200
AC = 14.14213… cm
Halve it to get AM. M is the midpoint of the diagonal.
AM = 14.14213… ÷ 2 = 7.07106… cm
AM² = 50
Use the right-angled triangle VMA. VM is the vertical height, so the angle at M is a right angle. VA is the hypotenuse.
VA² = AM² + VM² = 50 + 12² = 50 + 144 = 194
Take the square root and round to 1 d.p. We want VA, not VA².
VA = √194 = 13.92838…
VA = 13.9 cm

Check. The slant edge (13.9) is longer than the height (12) and longer than AM (7.1), as the hypotenuse should be. ✓

Watch out. AM is half the base diagonal, not half a side of the base.

Worked example 20.11.4

For the same pyramid, work out the angle between the edge VA and the base ABCD, correct to 1 decimal place.

Method. Find the shadow of the edge on the base (from the centre M to the corner), then use trigonometry in the right-angled triangle VMA.

Find the shadow of VA on the base. The angle between the edge VA and the base is the angle between VA and its shadow AM (M is the centre of the base). The angle we want is angle VAM.
Triangle VMA is right-angled at M.
Find AM. AM is half of the base diagonal AC, and we find AC using Pythagoras.
AC² = 10² + 10² = 200, so AC = 14.14213… cm
AM = 14.14213… ÷ 2 = 7.07106… cm
Label the sides for trigonometry. From angle VAM, the opposite side is VM = 12 cm and the adjacent side is AM = 7.07106… cm. That is TOA.
tan VAM = 127.07106… = 1.69705…
Use the inverse tangent and round. The inverse button turns the tangent back into an angle.
angle VAM = tan⁻¹(1.69705…) = 59.49104…°
angle VAM = 59.5°

Check. Use sine instead: VA = √(50 + 144) = 13.92838…, and sin VAM = 12 ÷ 13.92838… = 0.86154…; sin 59.5° = 0.862. ✓

Watch out. The right angle is at M (the centre of the base), not at the corner.

Worked example 20.11.5

For the same pyramid, work out the angle between the sloping face VAB and the base, correct to 1 decimal place.

Method. The angle between a sloping face and the base is found in the right-angled triangle formed by the height, the line from the centre M to the middle N of a base edge, and the line VN down the face.

Find the right-angled triangle. N is the midpoint of AB. MN is half of the base edge, and VM is the height. The angle at M is a right angle. The angle between the face VAB and the base is the angle VNM.
MN = 10 ÷ 2 = 5 cm, VM = 12 cm
Label from the angle VNM. VM is opposite the angle and MN is adjacent. That is TOA.
tan VNM = 125
Use the inverse tangent. Turn the ratio into an angle.
VNM = tan⁻¹(2.4) = 67.38013…°
Round.
Angle = 67.4°

Check. The slant height VN = √(5² + 12²) = 13, and sin VNM = 12 ÷ 13 = 0.92307…; sin 67.4° = 0.923. ✓

Watch out. Use the midpoint N of the base edge (half the side), not the corner A (half the diagonal).

Watch out

In 3D questions the picture is not drawn to scale, so do not measure from it. Draw the right-angled triangle separately, marking the right angle. And remember that the angle between a line and a plane is measured from the shadow of the line, not from an edge that is just nearby.

Chapter summary

Skill	Method
Hypotenuse (Pythagoras)	Add the squares of the two shorter sides, then square root: c = √(a² + b²)
Shorter side (Pythagoras)	Subtract the squares: a = √(c² − b²)
Is a triangle right-angled?	Does (short)² + (short)² = (longest)²? If equal, yes
Distance between two points	Horizontal and vertical steps are the shorter sides; d² = (x step)² + (y step)²
Isosceles and equilateral triangles	Split down the middle: half the base, the height, the slanting side make a right-angled triangle
Label the sides	Hypotenuse is opposite the right angle; opposite is across from θ; adjacent is next to θ
SOH CAH TOA	sin θ = O÷H, cos θ = A÷H, tan θ = O÷A
Find a side with trigonometry	Label, choose the ratio, write the equation, rearrange (unknown on top: multiply, on the bottom: divide), calculate (degree mode)
Find an angle with trigonometry	Write the ratio as a fraction, then use the inverse: θ = sin⁻¹, cos⁻¹ or tan⁻¹
Elevation and depression	Measured from the horizontal; the angle of depression from the top equals the angle of elevation from the bottom (alternate angles)
Bearings	Clockwise from north, three figures; back bearing = bearing ± 180°; east = d sin b, north = d cos b
Exact values	sin 30° = ½, sin 45° = √2÷2, sin 60° = √3÷2; cos is the reverse; tan 30° = √3÷3, tan 45° = 1, tan 60° = √3
Sine rule (Higher)	a÷sin A = b÷sin B = c÷sin C; needs a matching pair; check for the obtuse angle
Cosine rule (Higher)	a² = b² + c² − 2bc cos A; for SAS and SSS; cos A = (b² + c² − a²) ÷ 2bc
Area of a triangle (Higher)	Area = ½ ab sin C (C is between sides a and b)
3D problems (Higher)	Find right-angled triangles, draw them flat; angle between a line and a plane = angle with its shadow

Can you…?

I can use Pythagoras’ theorem to find the hypotenuse and a shorter side.
I can decide whether a triangle is right-angled and recognise Pythagorean triples.
I can solve problems with coordinates and with isosceles and equilateral triangles using Pythagoras.
I can label the opposite, adjacent and hypotenuse from any angle.
I can use sin, cos and tan to find a side, including when the unknown is on the bottom.
I can use sin⁻¹, cos⁻¹ and tan⁻¹ to find an angle.
I can solve problems with angles of elevation and depression and with bearings.
I know the exact values of sin, cos and tan for 0°, 30°, 45°, 60° and 90°.
Higher: I can use the sine rule, including the ambiguous case.
Higher: I can use the cosine rule to find a side or an angle.
Higher: I can find the area of a triangle with ½ ab sin C.
Higher: I can solve 3D problems and problems that combine several rules.

Key terms

Hypotenuse: The longest side of a right-angled triangle, opposite the right angle.
Pythagoras’ theorem: In a right-angled triangle, a² + b² = c², where c is the hypotenuse.
Pythagorean triple: Three whole numbers a, b, c with a² + b² = c², such as 3, 4, 5.
Opposite: The side across from the angle θ, not touching it.
Adjacent: The side next to the angle θ that is not the hypotenuse.
Sine, cosine, tangent: The ratios O÷H, A÷H and O÷A of a right-angled triangle.
Inverse function: sin⁻¹, cos⁻¹, tan⁻¹: turns a ratio back into an angle.
Angle of elevation: The angle upwards from the horizontal to a line of sight.
Angle of depression: The angle downwards from the horizontal to a line of sight.
Bearing: A direction measured clockwise from north, written with three figures.
Exact value: An answer left as a fraction or surd, not as a rounded decimal.
Rationalise: Remove a square root from the bottom of a fraction by multiplying top and bottom by it.
Sine rule: a÷sin A = b÷sin B = c÷sin C (Higher).
Cosine rule: a² = b² + c² − 2bc cos A (Higher).
Included angle: The angle between two given sides.
Space diagonal: A line joining opposite corners of a cuboid through the inside of the solid.
Projection (shadow): The shadow of a line on a plane, used to find the angle between a line and a plane.

Total 55 marks. Allow about 55 minutes.

← Chapter 19: Volume, surface area and 3D shapes Chapter 21: Transformations, congruence and similarity →

GCSE Maths Pythagoras and Trigonometry (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

You should already know

Where this leads

Exercise P Do you remember?

Pythagoras’ theorem: finding the hypotenuse and a shorter side

Key ideas

Watch out

Exercise 20A Pythagoras with whole-number answers

Exercise 20B Pythagoras with a calculator

Is it right-angled? Pythagorean triples and problems in context

Key ideas

Watch out

Exercise 20C Right-angled triangles in problems

Pythagoras with coordinates, isosceles and equilateral triangles

Key ideas

Watch out

Exercise 20D Coordinates, isosceles and equilateral triangles

Trigonometry: finding sides (SOH CAH TOA)

Key ideas

Watch out

Exercise 20E Labelling sides and choosing the ratio

Exercise 20F Finding sides with sin, cos and tan

Trigonometry: finding angles

Key ideas

Watch out

Exercise 20G Finding angles with the inverse functions

Angles of elevation and depression, and bearings

Key ideas

Watch out

Exercise 20H Elevation, depression and bearings

Exact values of sin, cos and tan

Key ideas

Watch out

Exercise 20I Exact values of trigonometric ratios

The sine rule

Key ideas

Watch out

Exercise 20J The sine rule

The cosine rule

Key ideas

Watch out

Exercise 20K The cosine rule

Area of a triangle: ½ab sin C

Key ideas

Watch out

Exercise 20L Area of a triangle using ½ab sin C

Trigonometry and Pythagoras in three dimensions, and combined problems

Key ideas

Watch out

Exercise 20M Trigonometry and Pythagoras in 3D

Exercise 20N Combined problems

Chapter summary

Can you…?

Key terms

Exercise 20Z Check-up (non-calculator)

Exercise 20X Exam-style questions