Websink Prep UKGCSE MathsTextbook contentsMaths textbooksYear 1Year 2Year 4Year 5Year 6GCSEBrowse allPast papersFind a tutorBecome a tutor
GCSE Maths · EdexcelChapter 20 · Pythagoras and trigonometry
Chapter 20 · Foundation tier

GCSE Maths Pythagoras and Trigonometry (Foundation): 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°

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, Foundation 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.

Exercise P  Do you remember?

Non-calculator
1
FWork out each of these.
(a)72
(b)122
(c)√81
(d)√144
2
FRound each number to 1 decimal place. (These matter because Pythagoras and trigonometry answers are often rounded.)
(a)7.8102
(b)12.0467
(c)0.5736
3
FSolve each equation.
(a)x ÷ 5 = 3.2
(b)12 ÷ x = 3
4
FThis triangle has a right angle and an angle of 35°. Work out the size of the angle marked x.
35°x
5
FWork out the area of this triangle.
10 cm6 cm
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).

shorter sideshorter sidehypotenuse
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.

3² = 94² = 165² = 25
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.
5 cm12 cmx

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.

  1. 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.
  2. 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
  3. Square each short side. Squaring means multiplying the number by itself.
    52 = 5 × 5 = 25
    122 = 12 × 12 = 144
  4. Add the two squares. This gives x2, the square of the hypotenuse.
    x2 = 25 + 144 = 169
  5. 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.
7 cm9 cmx

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.

  1. 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.
  2. 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
  3. Square each short side. Squaring means multiplying the number by itself.
    72 = 7 × 7 = 49
    92 = 9 × 9 = 81
  4. Add the two squares. This gives x2, the square of the hypotenuse.
    x2 = 49 + 81 = 130
  5. 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……
  6. 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.
x8 cm17 cm

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

  1. 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.
  2. 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
  3. Square the known numbers. Squaring means multiplying the number by itself.
    82 = 8 × 8 = 64
    172 = 17 × 17 = 289
    x2 + 64 = 289
  4. 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
  5. 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.
6.2 mx9.4 m

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

  1. 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.
  2. 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
  3. 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
  4. 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
  5. Take the square root. We want x, not x2.
    x = √49.92 = 7.06540…
  6. 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.

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.

Exercise 20A  Pythagoras with whole-number answers

Non-calculator
1
FWork out each square.
(a)42
(b)72
(c)92
(d)122
(e)152
(f)202
2
FWork out each square root.
(a)√25
(b)√49
(c)√81
(d)√144
(e)√169
(f)√225
3
FFor each right-angled triangle, write down the letters of the hypotenuse.
(a)
PQR
(b)
ABC
(c)
XYZ
4
FWork out the length of the hypotenuse, x.
3 cm4 cmx
5
FWork out the length of the hypotenuse, x, in each triangle. All the lengths are in centimetres.
(a)
6 cm8 cmx
(b)
5 cm12 cmx
(c)
8 cm15 cmx
(d)
20 cm21 cmx
6
FWork out the length of the hypotenuse, x.
24 cm7 cmx
7
FWork out the length of the shorter side, x, in each triangle. All the lengths are in centimetres.
(a)
x5 cm13 cm
(b)
6 cmx10 cm
(c)
x24 cm25 cm
(d)
15 cmx17 cm
8
FWork out the length of the shorter side, x.
x40 cm41 cm
9
PSA ladder 13 m long leans against a vertical wall. The foot of the ladder is 5 m from the wall on level ground. How high up the wall does the ladder reach?
5 mhladder 13 m
10
PSThe diagram shows a rectangle that is 24 cm long and 10 cm wide. Work out the length of the diagonal, x.
24 cm10 cmx
11
FWork out the length of the hypotenuse, x, in each triangle. Leave your answer as a square root (do not use a decimal). All lengths are in centimetres.
(a)
2 cm3 cmx
(b)
1 cm5 cmx
12
RAisha says: “The shorter sides are 6 cm and 8 cm, so the hypotenuse is 6 + 8 = 14 cm.” Explain why Aisha is wrong, and work out the correct length.
ABC6 cm8 cm14 cm
13
RKim says that the hypotenuse of this triangle is 9 cm. Without using Pythagoras’ theorem, explain why Kim must be wrong. Then work out the correct length, to 1 decimal place.
PQR10 cm6 cm
14
RA right-angled triangle has a hypotenuse of 25 cm and one shorter side of 7 cm. Show that its perimeter is 56 cm.
7 cmx25 cm

Exercise 20B  Pythagoras with a calculator

Calculator allowed
1
FWork out the length of the hypotenuse, x, in each triangle. Give your answers correct to 1 decimal place. All lengths are in centimetres.
(a)
4 cm7 cmx
(b)
6 cm9 cmx
(c)
8.5 cm5.2 cmx
(d)
12 cm12 cmx
2
FWork out the length of the shorter side, x, in each triangle. Give your answers correct to 1 decimal place. All lengths are in centimetres.
(a)
x5 cm9 cm
(b)
8.4 cmx14 cm
(c)
x6 cm7.5 cm
(d)
4.2 cmx15 cm
3
FWork out the length of the hypotenuse, x. Give your answer correct to 2 decimal places.
15.3 m8.7 mx
4
FWork out the length of the shorter side, x. Give your answer correct to 3 significant figures.
x19.4 mm23.6 mm
5
PSA television screen is a rectangle 80 cm wide and 45 cm high. Work out the length of the diagonal of the screen. Give your answer to the nearest centimetre, then say how many inches this is to the nearest inch (1 inch = 2.54 cm).
80 cm45 cmx
6
PSSam walks from one corner of a rectangular field to the opposite corner. The field is 120 m long and 90 m wide. He can walk along the two edges, or straight across the field along the diagonal. How many metres shorter is the diagonal route?
120 m90 md
7
PSA ship sails 15 km due north and then 8 km due east. How far is the ship from where it started?
15 km8 kmdStartNEnd
8
PSA vertical flagpole is 7.5 m tall. A wire runs from the top of the pole to a point on the ground 4 m from the foot of the pole. Work out the length of the wire.
4 m7.5 mx
9
PSA wheelchair ramp is 4.2 m long and rises 0.35 m. Work out the horizontal distance the ramp covers. Give your answer correct to 2 decimal places.
d0.35 mramp 4.2 m
10
RChallenge. A square has diagonals of length 10 cm. Use Pythagoras’ theorem to work out the length of one side of the square, correct to 1 decimal place. (Hint: the diagonal splits the square into two right-angled triangles whose two shorter sides are equal.)
ss10 cm
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.

ABC7 cm24 cm25 cm
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:

  1. Draw a sketch and mark the right angle.
  2. Mark the lengths you know, and call the one you want x.
  3. Decide whether x is the hypotenuse (the longest side, opposite the right angle) or a shorter side.
  4. Use Pythagoras: add the squares for the hypotenuse, subtract for a shorter side.
  5. 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.
7 cm24 cm25 cm

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

  1. 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.
  2. Square the two shorter sides and add. This is the left-hand side of Pythagoras’ theorem.
    72 + 242 = 49 + 576 = 625
  3. Square the longest side. This is the right-hand side.
    252 = 625
  4. 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.
6 cm7 cm10 cm

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

  1. 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.
  2. Square the two shorter sides and add. This is the left-hand side of Pythagoras’ theorem.
    62 + 72 = 36 + 49 = 85
  3. Square the longest side. This is the right-hand side.
    102 = 100
  4. 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.
1.4 mhladder 5.6 m

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

  1. 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.
  2. 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.
  3. 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
  4. 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
  5. 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
  6. Take the square root. We want h, not h2.
    h = √29.4 = 5.42217…
  7. 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.

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.

Exercise 20C  Right-angled triangles in problems

Calculator allowed
1
FIs each triangle right-angled? Show your working. All the lengths are in centimetres.
(a)
3 cm4 cm5 cm
(b)
5 cm12 cm13 cm
(c)
6 cm8 cm11 cm
(d)
9 cm12 cm15 cm
(e)
7 cm8 cm12 cm
(f)
20 cm21 cm29 cm
2
FEach triangle is right-angled and the lengths are in centimetres. Find the missing length x (a whole number).
(a)(3 whole numbers 8, 15, x)
8 cm15 cmx
(b)(9, x, 41)
x9 cm41 cm
(c)(x, 21, 29)
21 cmx29 cm
(d)(12, 16, x)
12 cm16 cmx
3
FA builder marks out the corner of a patio. She measures 0.9 m along one edge and 1.2 m along the other edge. The distance between the two marks is 1.5 m. Is the corner a right angle? You must show your working.
ABC0.9 m1.2 m1.5 m
4
F3, 4, 5 is a Pythagorean triple because 9 + 16 = 25.
(a)Show that 6, 8, 10 is a Pythagorean triple.
ABC6810
(b)Show that 30, 40, 50 is a Pythagorean triple.
ABC304050
5
PSA ladder reaches 5.2 m up a vertical wall. The foot of the ladder is 1.5 m from the wall. Work out the length of the ladder, correct to 2 decimal places.
1.5 m5.2 mL
6
PSA boat sails 24 km due west and then 10 km due south. Work out the distance of the boat from its starting point.
24 km10 kmdStartWEnd
7
PSThe diagonals of a rhombus are 16 cm and 12 cm long. Work out the length of one side of the rhombus, x.
x16 cm12 cm
8
PSTwo cyclists leave the same place at the same time. One cycles north at 12 km/h and the other cycles east at 5 km/h. How far apart are they after 2 hours?
24 km10 kmdStartC1C2
9
PSA joiner checks that a door frame is square. The frame is 0.9 m wide and 2.1 m tall, and a diagonal measures 2.3 m. Is the frame exactly square (right-angled)? Give a reason using Pythagoras’ theorem.
ABC0.9 m2.1 mdiagonal 2.3 m
10
PSA vertical pole is 6 m tall. A wire 10 m long runs from the top of the pole to a point on level ground. How far is that point from the foot of the pole?
dpole 6 mwire 10 m
11
RRaj says: “This triangle is not right-angled because 8 + 15 = 23, and 23 is not 16.” Explain why Raj’s method is wrong. Then use the correct method to decide whether the triangle is right-angled.
ABC8 cm15 cm16 cm
12
RChallenge. Jada says: “If I double every length of a right-angled triangle, the new triangle is still right-angled.” Use the triangle 5 cm, 12 cm, 13 cm to show that Jada is right for this triangle (the doubled triangle is drawn). Then explain why this must be true for any right-angled triangle.
ABC10 cm24 cm26 cm
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.

xy-11234567891011-11234567891011A(1, 2)B(7, 10)
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.

13 cm13 cm10 cmh
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.

ABC8 cm8 cm60°60°h4 cm4 cm
Equilateral triangle of side 8 cm: h2 + 42 = 82.

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).
xy-11234567891011-11234567891011A(1, 2)B(7, 10)

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

  1. 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.
  2. 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
  3. Use Pythagoras’ theorem. The distance d is the hypotenuse.
    d2 = 62 + 82 = 36 + 64 = 100
  4. 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.
xy-4-3-2-1123456-3-2-11234567P(−3, 4)Q(5, −2)

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

  1. 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.
  2. 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
  3. Use Pythagoras’ theorem. The distance d is the hypotenuse.
    d2 = 82 + 62 = 64 + 36 = 100
  4. 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.
13 cm13 cm10 cmh

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. An isosceles triangle has two equal sides.

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

  1. 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
  2. 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
  3. Square and subtract. We are finding a shorter side, so we subtract.
    h2 = 169 − 25 = 144
  4. Take the square root. The height is h, not h2.
    h = √144 = 12 cm
  5. 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
  6. 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.
ABC8 cm8 cm60°60°h4 cm4 cm

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. An isosceles triangle has two equal sides.

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

  1. 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
  2. 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
  3. Square and subtract. We are finding a shorter side, so we subtract.
    h2 = 64 − 16 = 48
  4. 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.

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.

Exercise 20D  Coordinates, isosceles and equilateral triangles

Calculator allowed
1
FWork out the distance between each pair of points. Give your answers correct to 1 decimal place when they are not whole numbers.
(a)A(1, 2) and B(4, 6)
xy-11234567-11234567A(1, 2)B(4, 6)
(b)C(0, 3) and D(5, −9)
xy24681012-10-8-6-4-224C(0, 3)D(5, −9)
(c)P(−4, 2) and Q(3, 5)
xy-5-4-3-2-11234-112345678P(−4, 2)Q(3, 5)
(d)R(−6, −1) and S(2, 4)
xy-7-6-5-4-3-2-1123-2-112345678R(−6, −1)S(2, 4)
2
PSOn a map, each grid square is 1 km wide. A town P is at (2, 3) and a town Q is at (14, 8). Work out the straight-line distance from P to Q.
xy24681012142468101214P(2, 3)Q(14, 8)
3
RA(2, 1), B(6, 4) and C(2, 7) are the vertices of a triangle. Show that triangle ABC is isosceles.
xy-112345678-112345678A(2, 1)B(6, 4)C(2, 7)
4
FEach isosceles triangle has the equal sides and the base shown. Work out the perpendicular height h of each triangle.
(a)Equal sides 10 cm, base 12 cm
10 cm10 cm12 cmh
(b)Equal sides 13 cm, base 10 cm
13 cm13 cm10 cmh
(c)Equal sides 17 cm, base 16 cm
17 cm17 cm16 cmh
(d)Equal sides 6.5 cm, base 5 cm
6.5 cm6.5 cm5 cmh
5
FAn isosceles triangle has two sides of length 10 cm and a base of 12 cm. Work out its area.
10 cm10 cm12 cmh
6
FAn equilateral triangle has sides of length 10 cm. Work out its height. Give your answer correct to 2 decimal places.
ABC10 cm10 cm60°60°h5 cm5 cm
7
PSA roof is supported by an isosceles triangular frame. The two sloping beams are each 5.2 m long and the horizontal span is 9.6 m. Work out the height of the roof frame.
5.2 m5.2 m9.6 mh
8
PSThe cross-section of a tent is an isosceles triangle. The sloping sides are 1.9 m long and the base is 2.4 m wide. Work out the height of the tent. Give your answer correct to 2 decimal places.
1.9 m1.9 m2.4 mh
9
PSA road sign is an equilateral triangle of side 50 cm. Work out the area of the sign. Give your answer correct to 1 decimal place.
ABC50 cm50 cm60°60°h25 cm25 cm
10
RZara says: “The distance between A(1, 2) and B(4, 6) is 3 + 4 = 7.” Explain why Zara is wrong and work out the correct distance.
xy-11234567-11234567A(1, 2)B(4, 6)
11
RIsosceles triangle ABC has AB = AC = 17 cm and BC = 16 cm. Show that its area is 120 cm2.
17 cm17 cm16 cmh
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.

36241230°
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”).

adjacentoppositehypotenuseθ
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.

OSHsin θACHcos θOTAtan θ
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.

  1. Label the sides O, A, H from the angle θ.
  2. Choose the ratio: look at the side you know and the side you want, and find the two letters in SOH CAH TOA.
  3. Write the ratio with the numbers: sin θ = O ÷ H becomes, for example, sin 35° = x ÷ 8.2.
  4. 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.
  5. 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.
x8.2 cm35°

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.

  1. 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.
  2. 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)
  3. Put the numbers into the ratio. Replace θ with the angle and the sides with the lengths.
    sin 35° = x8.2
  4. 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°
  5. Use the calculator (degree mode). First find sin 35°, then finish the calculation.
    sin 35° = 0.57357…
    x = 8.2 × 0.57357… = 4.70332…
  6. 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.
12 mx40°

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.

  1. 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.
  2. 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)
  3. Put the numbers into the ratio. Replace θ with the angle and the sides with the lengths.
    cos 40° = 12x
  4. 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
  5. Divide both sides by cos 40°. This leaves x on its own.
    x = 12cos 40° = 12 ÷ cos 40°
  6. Use the calculator (degree mode). First find cos 40°, then finish the calculation.
    cos 40° = 0.76604…
    x = 12 ÷ 0.76604… = 15.66488…
  7. 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.
7.5 cmx52°

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.

  1. 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.
  2. 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)
  3. Put the numbers into the ratio. Replace θ with the angle and the sides with the lengths.
    tan 52° = x7.5
  4. 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°
  5. Use the calculator (degree mode). First find tan 52°, then finish the calculation.
    tan 52° = 1.27994…
    x = 7.5 × 1.27994… = 9.59956…
  6. 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.
x14.2 mm27°

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.

  1. 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.
  2. 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)
  3. Put the numbers into the ratio. Replace θ with the angle and the sides with the lengths.
    tan 27° = 14.2x
  4. 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
  5. Divide both sides by tan 27°. This leaves x on its own.
    x = 14.2tan 27° = 14.2 ÷ tan 27°
  6. Use the calculator (degree mode). First find tan 27°, then finish the calculation.
    tan 27° = 0.50952…
    x = 14.2 ÷ 0.50952… = 27.86906…
  7. 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.

Exercise 20E  Labelling sides and choosing the ratio

Non-calculator
1
FThe angle θ is marked in each right-angled triangle. Name the hypotenuse, the opposite side and the adjacent side.
(a)
ABCθ
(b)
PQRθ
(c)
XYZθ
2
FWrite down what each of S, O, H, C, A, T stands for, and then complete each ratio.
OSHsin θACHcos θOTAtan θ
(a)sin θ =
(b)cos θ =
(c)tan θ =
3
FWhich ratio, sin, cos or tan, would you use to find x? Write down the ratio only.
(a)
x8 cm35°
(b)
12 mx40°
(c)
x6.4 cm55°
(d)
x15 mm28°
4
FWrite down an equation that you could use to find x. Do not solve it.
(a)
x8 cm35°
(b)
12 mx40°
(c)
x6.4 cm55°
(d)
x9.5 cm62°
5
FFor the triangle shown, write down sin θ, cos θ and tan θ as fractions.
12 cm5 cm13 cmθ
6
FThe triangle ABC has a right angle at C. For angle A, write down the opposite and adjacent sides. Then do the same for angle B. What do you notice?
CBA4 cm3 cm5 cmBA
7
PSFor each situation, choose the ratio you would use. Label the sides of the sketch first.
(a)A 4 m ladder makes an angle of 70° with the ground. You want its height h up the wall. Which ratio (sin, cos or tan)?
groundhladder 4 m70°
(b)A ramp rises 0.5 m and makes an angle of 12° with the ground. You want the horizontal distance d. Which ratio?
drise 0.5 m12°
(c)A kite string is 30 m long and makes an angle of 55° with the level ground. You want the horizontal distance d from the kite to the person. Which ratio?
dstring 30 m55°
8
RDev says that tan x = 513 for this triangle. Explain his mistake and write down the correct value of tan x as a fraction.
12 cm5 cm13 cmx
9
RExplain why sin θ and cos θ can never be bigger than 1 for an angle θ in a right-angled triangle.
AOH (longest)θ
10
PSThe diagram shows three right-angled triangles that share a 30° angle. Work out (opposite) ÷ (hypotenuse) for each triangle and say what you notice.
36241230°

Exercise 20F  Finding sides with sin, cos and tan

Calculator allowed
1
FUse a calculator in degree mode to work out each value. Give each answer correct to 3 decimal places.
(a)sin 40°
40°
(b)cos 25°
25°
(c)tan 63°
63°
(d)sin 72°
72°
(e)cos 58°
58°
(f)tan 18°
18°
2
FWork out the length of the side marked x in each triangle. Give your answers correct to 1 decimal place. All the lengths are in centimetres.
(a)
x10 cm30°
(b)
x7.5 cm42°
(c)
x12 cm65°
(d)
x9.6 cm23°
3
FWork out the length of the side marked x in each triangle. Give your answers correct to 1 decimal place. All the lengths are in centimetres.
(a)
x15 cm40°
(b)
x8.2 cm55°
(c)
x11 cm28°
(d)
x6.5 cm71°
4
FWork out the length of the side marked x in each triangle. Give your answers correct to 1 decimal place. All the lengths are in centimetres.
(a)
8 cmx50°
(b)
12.5 cmx33°
(c)
5.4 cmx68°
(d)
15 cmx20°
5
FWork out the length of the side marked x in each triangle. Be careful: in some of them x will be on the bottom of the fraction. Give your answers correct to 1 decimal place. All the lengths are in centimetres.
(a)
9 cmx35°
(b)
7 cmx48°
(c)
x5.5 cm62°
(d)
14 cmx26°
6
FWork out the length x. Give your answer correct to 3 significant figures.
x19.4 cm36°
7
FWork out the length x. Give your answer correct to 2 decimal places.
3.8 mx57°
8
PSA ladder 6.5 m long leans against a vertical wall. The ladder makes an angle of 68° with the horizontal ground. How high up the wall does the ladder reach? Give your answer correct to 2 decimal places.
groundhladder 6.5 m68°
9
PSA slide is 3.2 m long and slopes at 38° to the horizontal. Work out the vertical height of the top of the slide above the ground. Give your answer correct to 2 decimal places.
hslide 3.2 m38°
10
PSA kite is flying on a string 45 m long. The string makes an angle of 52° with the level ground. Work out the horizontal distance from the person to the point on the ground below the kite. Give your answer correct to 1 decimal place.
dstring 45 m52°
11
PSA wheelchair ramp rises 0.6 m and makes an angle of 7° with the ground. How long is the ramp? Give your answer correct to 2 decimal places.
rise 0.6 mL7°
12
PSIn this right-angled triangle the hypotenuse is 12 cm and one angle is 34°. Find x and y, correct to 1 decimal place.
xy12 cm34°
(a)y (opposite the 34° angle)
(b)x (next to the 34° angle)
13
RAnya works out x in this triangle as 10 × cos 40° = 7.7 cm. Explain her mistake and find the correct value of x, to 1 decimal place.
x10 cm40°
14
RChallenge. A right-angled triangle has a hypotenuse of 10 cm and an angle of 36°. Work out the perimeter of the triangle, correct to 1 decimal place.
??10 cm36°
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.

6.7 cm10 cmθ
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.

  1. Label the two sides you know as O, A or H, measured from the angle you want.
  2. Choose the ratio (SOH, CAH or TOA) that uses those two sides.
  3. Write the ratio as a fraction with the numbers: for example tan θ = 5 ÷ 8.
  4. Use the inverse button on the calculator: θ = tan−1(5 ÷ 8).
  5. 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.
6.7 cm10 cmx

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 as a fraction, then use the inverse function to turn the ratio into an angle.

  1. 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.
  2. 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)
  3. Put the numbers into the ratio. Write the sides as a fraction (top = opposite, bottom = hypotenuse).
    sin x = 6.710 = 0.67
  4. 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…°
  5. 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.
8 cm5 cmx

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 as a fraction, then use the inverse function to turn the ratio into an angle.

  1. 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.
  2. 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)
  3. Put the numbers into the ratio. Write the sides as a fraction (top = opposite, bottom = adjacent).
    tan x = 58 = 0.625
  4. 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…°
  5. 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.
4.5 m9 mx

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 as a fraction, then use the inverse function to turn the ratio into an angle.

  1. 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.
  2. 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)
  3. Put the numbers into the ratio. Write the sides as a fraction (top = adjacent, bottom = hypotenuse).
    cos x = 4.59 = 0.5
  4. 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°
  5. 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.
10 cm10 cm12 cmx

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 as a fraction, then use the inverse function to turn the ratio into an angle.

  1. 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).
  2. 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.
  3. 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)
  4. Put the numbers into the ratio. Write the sides as a fraction (top = adjacent, bottom = hypotenuse).
    cos x = 610 = 0.6
  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(610) = cos−1(0.6) = 53.13010…°
  6. 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.

Exercise 20G  Finding angles with the inverse functions

Calculator allowed
1
FUse the inverse function on your calculator to find x. Give each answer correct to 1 decimal place.
(a)sin x = 0.6
x
(b)cos x = 0.25
x
(c)tan x = 1.2
x
(d)sin x = 37
x
2
FWork out the size of angle x in each triangle. Give your answers correct to 1 decimal place. All lengths are in centimetres.
(a)
4 cm9 cmx
(b)
7.5 cm12 cmx
(c)
3.2 cm5 cmx
(d)
16 cm21 cmx
3
FWork out the size of angle x in each triangle. Give your answers correct to 1 decimal place. All lengths are in centimetres.
(a)
6 cm10 cmx
(b)
3.8 cm8 cmx
(c)
12 cm15 cmx
(d)
5.5 cm7.2 cmx
4
FWork out the size of angle x in each triangle. Give your answers correct to 1 decimal place. All lengths are in centimetres.
(a)
8 cm5 cmx
(b)
4 cm9 cmx
(c)
7.5 cm7.5 cmx
(d)
11 cm2.4 cmx
5
FWork out the size of angle x. Give your answer correct to 1 decimal place.
12 cm5 cmx
6
PSA ramp is 5 m long and rises 0.8 m. Work out the angle the ramp makes with the horizontal, correct to 1 decimal place.
0.8 m5 mx
7
PSA ladder 6 m long stands on level ground with its foot 1.6 m from a vertical wall. Work out the angle between the ladder and the ground, correct to 1 decimal place.
1.6 m6 mx
8
PSThe diagram shows a rectangle 12 cm long and 5 cm wide with a diagonal drawn in. Work out the angle x between the diagonal and the longer side, correct to 1 decimal place.
12 cm5 cm
9
PSA road has a gradient of 1 in 8. This means it rises 1 m for every 8 m measured horizontally. Work out the angle the road makes with the horizontal, correct to 1 decimal place.
8 m1 mx
10
PSThe line OP joins the origin O to the point P(8, 5). Work out the angle that OP makes with the x-axis, correct to 1 decimal place.
xy-112345678910-11234567P(8, 5)x
11
RAli writes sin x = 1.3 for a right-angled triangle. Explain why this cannot be right.
OHx
12
RA right-angled triangle has a hypotenuse of 12 cm and a shorter side of 6 cm. Show that the angle opposite the 6 cm side is 30°.
6 cm12 cmx
13
RChallenge. Triangle ABC has a right angle at C, AC = 12 cm and BC = 5 cm. Work out the size of angle A and angle B, correct to 1 decimal place, and show that your three angles add up to 180°.
CAB12 cm5 cmAB
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.

θ60 m45 mTO
From O, the angle of elevation of the top T of the tower is θ. tan θ = 45 ÷ 60.
25°25°x80 mOB
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.

N 000°E 090°S 180°W 270°NE 045°SE 135°SW 225°NW 315°
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).

N060°20 kmPQ
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.
θ60 m45 mTO

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 as a fraction, then use the inverse function to turn the ratio into an angle.

  1. 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.
  2. 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.
  3. 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)
  4. Put the numbers into the ratio. Write the sides as a fraction (top = opposite, bottom = adjacent).
    tan x = 4560 = 0.75
  5. 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…°
  6. 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.
32°50 mhTO

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.

  1. 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.
  2. 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.
  3. 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)
  4. Put the numbers into the ratio. Replace θ with the angle and the sides with the lengths.
    tan 32° = h50
  5. 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°
  6. Use the calculator (degree mode). First find tan 32°, then finish the calculation.
    tan 32° = 0.62486…
    h = 50 × 0.62486… = 31.24346…
  7. 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.
25°25°x80 mOB

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.

  1. 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.
  2. 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.
  3. 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)
  4. Put the numbers into the ratio. Replace θ with the angle and the sides with the lengths.
    tan 25° = 80x
  5. 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
  6. Divide both sides by tan 25°. This leaves x on its own.
    x = 80tan 25° = 80 ÷ tan 25°
  7. Use the calculator (degree mode). First find tan 25°, then finish the calculation.
    tan 25° = 0.46630…
    x = 80 ÷ 0.46630… = 171.56055…
  8. 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.
N090°N12 km5 kmACABC

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. 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.

  1. 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.
  2. Find the distance AC. AC is the hypotenuse of the right-angled triangle ABC.
    AC2 = 122 + 52 = 144 + 25 = 169, so AC = 13 km
  3. 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…°
  4. 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°
  5. 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.
N060°20 kmPQ

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. 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.

  1. 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°.
  2. 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
  3. Find the distance east. H and O make SOH.
    East = 20 × sin 60° = 20 × 0.86602… = 17.32050… km = 17.3 km
  4. 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.

Exercise 20H  Elevation, depression and bearings

Calculator allowed
1
FLook at each diagram. Is the angle marked a an angle of elevation or an angle of depression? Give a reason.
(a)
aTO
(b)
aOB
2
FThe angle of elevation of the top of a flagpole from a point on level ground 30 m from its foot is 40°. Work out the height of the flagpole, correct to 1 decimal place.
40°30 mhTO
3
FA tree is 12 m tall. Ben stands 20 m from the base of the tree on level ground. Work out the angle of elevation of the top of the tree from Ben, correct to 1 decimal place.
x20 m12 mTBen
4
FFrom the top of a vertical lighthouse 60 m tall, the angle of depression of a boat is 15°. How far is the boat from the foot of the lighthouse? Give your answer correct to 1 decimal place.
15°15°d60 mOBoat
5
PSAn aircraft is flying at a height of 3000 m. From the airport the angle of elevation of the aircraft is 22°. Work out the horizontal distance of the aircraft from the airport, to the nearest metre.
22°d3000 mAircraftAirport
6
PSFrom a window 12 m above level ground, the angle of depression of a car is 35°. How far is the car from the wall of the building? Give your answer correct to 1 decimal place.
35°35°d12 mWindowCar
7
PSA ship leaves a port and sails 5 km east and then 12 km north. Work out the distance of the ship from the port, and the bearing of the ship from the port to the nearest degree.
N!x5 km12 kmdPortBShip
8
PSAn aircraft flies 150 km on a bearing of 040°. Work out how far (a) east and (b) north of its starting point the aircraft is. Give your answers correct to 1 decimal place.
N040°150 kmStartAircraft
(a)east
(b)north
9
PSA port P has a lighthouse A 8 km due north of it. A buoy B is 6 km due east of P. Work out (a) the distance AB, (b) the bearing of B from A, to the nearest degree.
N!x8 km6 kmABPAB
10
FWork out each back bearing.
(a)The bearing of B from A is 070°. Find the bearing of A from B.
N070°N!?AB
(b)The bearing of B from A is 125°. Find the bearing of A from B.
N125°N!?AB
(c)The bearing of B from A is 310°. Find the bearing of A from B.
N310°N!?AB
(d)The bearing of B from A is 205°. Find the bearing of A from B.
N205°N!?AB
11
RFrom the top A of a cliff, the angle of depression of a boat B is 30°. Explain why the angle of elevation of A from B is also 30°.
30°30°AB
12
RA hiker walks 6 km on a bearing of 035°, then 8 km on a bearing of 125°. Show that the hiker finishes 10 km from the start.
N035°N125°6 km8 kmACABC
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.

11√245°45°
11√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.

ABC2260°60°√311
1√3260°30°
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:

θ0°30°45°60°90°
sin θ012√22√321
cos θ1√32√22120
tan θ0√331√3undefined

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°.
1√3260°30°

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.

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

  1. 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).
  2. Write the ratio for sin. SOH: sin = opposite ÷ hypotenuse.
    sin 60° = √32
  3. 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.
1√3260°30°

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.

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

  1. 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).
  2. Write the ratio for tan. TOA: tan = opposite ÷ adjacent.
    tan 30° = 1√3
  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°.

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.

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

  1. Write down each exact value. Use the table or the special triangles. Do not use decimals.
    sin 60° = √32
    cos 30° = √32
  2. Put them into the calculation. Replace each trig value by its exact value.
    sin 60° × cos 30° = √32 × √32
  3. 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.
x10 cm60°

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.

  1. 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)
  2. Put the numbers in. The angle is one of the special angles, so use its exact value.
    sin 60° = x10
    sin 60° = √32
  3. 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.
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.

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.

  1. Write the equation. We know the exact value of cos x.
    cos x = √22
  2. 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
  3. 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.

Exercise 20I  Exact values of trigonometric ratios

Non-calculator
1
FWrite down the exact value of each of these. Do not use a calculator.
(a)sin 30°
30°
(b)cos 60°
60°
(c)tan 45°
45°
(d)sin 90°
(e)cos 0°
(f)tan 0°
2
FWrite down the exact value of each of these. Use the special triangles to help if you need to. Give fractions with rational denominators.
(a)sin 60°
1√3260°30°
(b)cos 30°
1√3260°30°
(c)sin 45°
11√245°45°
(d)tan 60°
1√3260°30°
(e)cos 45°
11√245°45°
(f)tan 30°
1√3260°30°
3
FUse the right-angled triangles in the diagram to show that tan 60° = √3.
1√3260°30°
4
FWork out the size of the acute angle x in each case.
(a)sin x = 12
x
(b)cos x = √22
x
(c)tan x = √3
x
(d)tan x = √33
x
5
FWork out the exact value of each expression. Show your working.
(a)sin 30° + cos 60°
1√3260°30°
(b)cos 45° × sin 45°
11√245°45°
(c)tan 45° − sin 30°
1√3260°30°
(d)sin 60° ÷ cos 60°
1√3260°30°
6
PSA right-angled triangle has a hypotenuse of 8 cm and a 30° angle. Work out the exact lengths of the other two sides.
(a)The side x (opposite the 30° angle)
x8 cm30°
(b)The side y (next to the 30° angle)
y8 cm30°
7
PSA ladder 6 m long leans against a vertical wall and makes an angle of 60° with the horizontal ground. Work out the exact height that the ladder reaches up the wall.
groundhladder 6 m60°
8
PSA right-angled triangle has a 45° angle. The side opposite the 45° angle is 4 cm. Work out the exact length of the hypotenuse.
4 cmx45°
9
REve says that sin 60° is double sin 30°, because 60 is double 30. Use exact values to show that Eve is wrong.
1√3260°30°
10
RIs it true that sin 45° + cos 45° = sin 90°? Show your working.
11√245°45°
11
RChallenge. Show that sin 60° = 2 sin 30° cos 30°.
1√3260°30°

Chapter summary

SkillMethod
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 pointsHorizontal and vertical steps are the shorter sides; d2 = (x step)2 + (y step)2
Isosceles and equilateral trianglesSplit down the middle: half the base, the height, the slanting side make a right-angled triangle
Label the sidesHypotenuse is opposite the right angle; opposite is across from θ; adjacent is next to θ
SOH CAH TOAsin θ = O÷H, cos θ = A÷H, tan θ = O÷A
Find a side with trigonometryLabel, choose the ratio, write the equation, rearrange (unknown on top: multiply, on the bottom: divide), calculate (degree mode)
Find an angle with trigonometryWrite the ratio as a fraction, then use the inverse: θ = sin−1, cos−1 or tan−1
Elevation and depressionMeasured from the horizontal; the angle of depression from the top equals the angle of elevation from the bottom (alternate angles)
BearingsClockwise from north, three figures; back bearing = bearing ± 180°; east = d sin b, north = d cos b
Exact valuessin 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.

Exercise 20Z  Check-up (non-calculator)

Non-calculator
1
Work out the length of the hypotenuse, x.
6 cm8 cmx
2
Work out the length of the shorter side, x.
x12 cm13 cm
3
Is a triangle with sides 9 cm, 12 cm and 15 cm right-angled? Show your working.
9 cm12 cm15 cm
4
Work out the distance between A(1, 1) and B(7, 9).
xy-112345678910-112345678910A(1, 1)B(7, 9)
5
An isosceles triangle has equal sides of 17 cm and a base of 16 cm. Work out its area.
17 cm17 cm16 cmh
6
Write down the hypotenuse, the side opposite θ and the side adjacent to θ.
ABCθ
7
Write down an equation to find x. Do not solve it.
7 cmx48°
8
Work out the exact value of sin 30° + cos 60°.
1√3260°30°
9
The bearing of B from A is 065°. Work out the bearing of A from B.
N065°N!?AB
10
Write down the exact values of tan 45° and sin 60°.
11√245°45°
(a)tan 45°
(b)sin 60°

Total 38 marks. Allow about 38 minutes.

Exercise 20X  Exam-style questions

Non-calculator
1
FWork out the length of x. Give your answer correct to 1 decimal place.[2 marks]
6.5 cm9.2 cmx
2
PSA ladder of length 7.5 m rests against a vertical wall with its foot on level ground, 2.1 m from the wall. Work out how high up the wall the ladder reaches.[3 marks]
2.1 mhladder 7.5 m
3
RA triangle has sides of length 11 cm, 60 cm and 61 cm. Is the triangle right-angled? You must show how you decide.[3 marks]
11 cm60 cm61 cm
4
PSA is the point (−2, 3) and B is the point (4, −5). Work out the length of AB.[3 marks]
xy-3-2-11234567-6-5-4-3-2-11234A(−2, 3)B(4, −5)
5
PSAn isosceles triangle has equal sides of 25 cm and a base of 14 cm. Work out the area of the triangle.[3 marks]
25 cm25 cm14 cmh
6
FWork out the length of x. Give your answer correct to 1 decimal place.[2 marks]
x14.5 cm28°
7
FWork out the size of angle x. Give your answer correct to 1 decimal place.[3 marks]
13.5 cm9.2 cmx
8
PSFrom a point 40 m from the foot of a vertical building on level ground, the angle of elevation of the top of the building is 35°. (a) Work out the height of the building, correct to 1 decimal place. (b) Imogen stands 100 m from the foot of the building. Work out the angle of elevation of the top of the building from Imogen, correct to 1 decimal place.[4 marks]
hTImogen
(a)(a)
(b)(b)
9
PSA boat leaves a harbour H and travels 8 km due north to a buoy B, then 15 km due east to a lighthouse L. Work out (a) the distance HL, (b) the bearing of L from H, to the nearest degree.[4 marks]
N!x8 km15 kmHLHBL
10
FWork out the exact value of sin 60° × tan 30°. Give your answer as a fraction.[2 marks]
1√3260°30°
11
PSA vertical pole is 9.6 m tall. A wire runs from the top of the pole to a peg in the ground. The wire makes an angle of 40° with the ground. Work out (a) the length of the wire, (b) the distance from the foot of the pole to the peg. Give your answers correct to 1 decimal place.[4 marks]
dpole 9.6 mL40°
(a)(a)
(b)(b)
12
PSThe diagram shows the end of a roof. It is an isosceles triangle with a horizontal base of 9 m. Each base angle is 35°. Work out (a) the height of the roof, (b) the length of one sloping rafter. Give your answers correct to 2 decimal places.[5 marks]
9 mrr35°35°
(a)(a)
(b)(b)
← Chapter 19: Volume, surface area and 3D shapesChapter 21: Transformations, congruence and similarity →

Websink Prep UK is an independent site. It is not affiliated with, endorsed by or sponsored by AQA, Pearson Edexcel, OCR, WJEC, Eduqas, CCEA, SQA, GL Assessment, CEM or the Department for Education. All exam board names and trademarks belong to their owners. Past papers are linked from the exam boards' own websites and are not hosted here.

About Websink Prep UK · Find a tutor · Privacy · Terms · Cookie settings