GCSE Maths Pythagoras and Trigonometry (Higher): check-up and exam-style questions

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: 6 cm and 8 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
   6² + 8² = x²
3. Square each short side. Squaring means multiplying the number by itself.
   6² = 6 × 6 = 36
   8² = 8 × 8 = 64
4. Add the two squares. This gives x², the square of the hypotenuse.
   x² = 36 + 64 = 100
5. Take the square root. We want x, not x². The square root undoes the squaring.
   x = √100 = 10 cm

Check. The hypotenuse must be the longest side: 10 is bigger than 6 and 8. Also 6² + 8² = 100 = 10². ✓

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

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 = 13 cm. The other known side = 12 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 + 12² = 13²
3. Square the known numbers. Squaring means multiplying the number by itself.
   12² = 12 × 12 = 144
   13² = 13 × 13 = 169
   x² + 144 = 169
4. Subtract 144 from both sides. We want x² on its own. Taking 144 away from both sides keeps the equation balanced.
   x² = 169 − 144 = 25
5. Take the square root. We want x, not x².
   x = √25 = 5 cm

Check. Put it back: 5² + 12² = 25 + 144 = 169 = 13². ✓ And 5 is shorter than the hypotenuse 13.

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.

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: 9, 12, 15. Longest = 15 cm.
2. Square the two shorter sides and add. This is the left-hand side of Pythagoras’ theorem.
   9² + 12² = 81 + 144 = 225
3. Square the longest side. This is the right-hand side.
   15² = 225
4. Compare the two answers. If they are equal, Pythagoras’ theorem works, so the triangle is right-angled. If not, it is not.
   225 = 225. They are equal, so the triangle is right-angled.

Check. Sides that fit Pythagoras’ theorem are called a Pythagorean triple. Here 3, 4, 5 is a smaller triple and every side has been multiplied by 3. ✓

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.

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: 9 − 1 = 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.

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 = 16 ÷ 2 = 8 cm
2. Pick one right-angled triangle. Its hypotenuse is the slanting side, 17 cm. Its shorter sides are the half-base and the height h.
   h² + 8² = 17²
3. Square and subtract. We are finding a shorter side, so we subtract.
   h² = 289 − 64 = 225
4. Take the square root. The height is h, not h².
   h = √225 = 15 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 × 16 × 15 = 120
6. Write the area with its units. Area is measured in square units. The answer is exact, so no rounding is needed.
   Area = 120 cm²

Check. Another way: two small triangles each have area ½ × 8 × 15 = 60.00; doubling gives 120.00, matching 120. ✓

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

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. Stand at the angle θ. The hypotenuse is opposite the right angle. The opposite side is across from θ. The adjacent side is the remaining one, next to θ.

1. Find the right angle and the hypotenuse. The right angle is marked with a square. The hypotenuse is the side that does not touch it.
   Right angle at A, so the hypotenuse is BC.
2. Find the opposite side. θ is at B. The opposite side is across the triangle from B, so it does not touch B.
   The side that does not touch B is AC, so the opposite is AC.
3. Find the adjacent side. The adjacent side is next to θ (it touches B) and it is not the hypotenuse.
   The side left over that touches B is AB, so the adjacent is AB.

Check. Every side has exactly one name. The three names are different. ✓

Watch out. Opposite and adjacent change if you change the angle. The hypotenuse never changes.

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, choose the ratio, then write it with the numbers. We do not solve it here.

1. Label the sides. Opposite: across from the angle. Adjacent: next to it. Hypotenuse: opposite the right angle.
   7 cm is the adjacent. x is the hypotenuse.
2. Choose the ratio. The two sides used decide it.
   CAH stands for Cos, Adjacent, Hypotenuse:
   cos θ = AH   (A = adjacent, H = hypotenuse)
3. Write the equation. Replace θ with 48°, A with the adjacent and H with the hypotenuse.
   cos 48° = 7x

Check. The side on the top of the fraction is the one named first in SOH, CAH or TOA. ✓

Watch out. The unknown can be on the top or the bottom of the fraction. Put each side in its own place.

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 30° = 12
   cos 60° = 12
2. Put them into the calculation. Replace each trig value by its exact value.
   sin 30° + cos 60° = 12 + 12
3. Work it out exactly. To add fractions, give them the same denominator, then add the tops.
   12 + 12 = 1

Check. Decimal check: 0.5 + 0.5 = 1, and 1 = 1. ✓

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.

Key words. A bearing is measured clockwise from north. The back bearing is the bearing for the return journey.

Method. The back bearing differs by 180°. Add 180° because 65° is less than 180°.

1. Decide whether to add or subtract. Bearings are between 000° and 360°.
   65° is less than 180°, so add 180°.
2. Work it out. Write the answer with three figures.
   065° + 180° = 245°

Check. 245 − 65 = 180. ✓

Watch out. Draw a north line at each point.

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: 6.5 cm and 9.2 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
   6.5² + 9.2² = x²
3. Square each short side. Squaring means multiplying the number by itself.
   6.5² = 6.5 × 6.5 = 42.25
   9.2² = 9.2 × 9.2 = 84.64
4. Add the two squares. This gives x², the square of the hypotenuse.
   x² = 42.25 + 84.64 = 126.89
5. Take the square root. We want x, not x². The square root undoes the squaring. Use the √ button on the calculator.
   x = √126.89 = 11.26454……
6. Round to 1 d.p.. The question tells us how accurate to be.
   x = 11.3 cm

Check. Square the answer: 11.3² = 127.69, which is close to 126.89. The hypotenuse 11.3 is longer than both 6.5 and 9.2. ✓

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

Marks. 1 mark for 6.5² + 9.2² = 126.89, 1 mark for the answer 11.3.

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 = 7.5 m. The other known side = 2.1 m. The side we want is h.
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
   h² + 2.1² = 7.5²
3. Square the known numbers. Squaring means multiplying the number by itself.
   2.1² = 2.1 × 2.1 = 4.41
   7.5² = 7.5 × 7.5 = 56.25
   h² + 4.41 = 56.25
4. Subtract 4.41 from both sides. We want h² on its own. Taking 4.41 away from both sides keeps the equation balanced.
   h² = 56.25 − 4.41 = 51.84
5. Take the square root. We want h, not h².
   h = √51.84 = 7.2 m

Check. Put it back: 7.2² + 2.1² = 51.84 + 4.41 = 56.25 = 7.5². ✓ And 7.2 is shorter than the hypotenuse 7.5.

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.

Marks. 1 mark for squaring both lengths, 1 mark for subtracting (56.25 − 4.41), 1 mark for the answer 7.2 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. 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: 11, 60, 61. Longest = 61 cm.
2. Square the two shorter sides and add. This is the left-hand side of Pythagoras’ theorem.
   11² + 60² = 121 + 3600 = 3721
3. Square the longest side. This is the right-hand side.
   61² = 3721
4. Compare the two answers. If they are equal, Pythagoras’ theorem works, so the triangle is right-angled. If not, it is not.
   3721 = 3721. 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.

Marks. 1 mark for 11² + 60², 1 mark for 61², 1 mark for comparing and stating the conclusion.

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: 4 − (−2) = 6, so the step is 6
   Vertical: −5 − 3 = −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.

Marks. 1 mark for the steps 6 and 8, 1 mark for 36 + 64, 1 mark for the answer 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. 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 = 14 ÷ 2 = 7 cm
2. Pick one right-angled triangle. Its hypotenuse is the slanting side, 25 cm. Its shorter sides are the half-base and the height h.
   h² + 7² = 25²
3. Square and subtract. We are finding a shorter side, so we subtract.
   h² = 625 − 49 = 576
4. Take the square root. The height is h, not h².
   h = √576 = 24 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 × 14 × 24 = 168
6. Write the area with its units. Area is measured in square units. The answer is exact, so no rounding is needed.
   Area = 168 cm²

Check. Another way: two small triangles each have area ½ × 7 × 24 = 84.00; doubling gives 168.00, matching 168. ✓

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

Marks. 1 mark for using half the base 7, 1 mark for the height 24, 1 mark for the area 168.

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.
   14.5 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 28° = x14.5
4. Multiply both sides by 14.5. x is being divided by 14.5. Multiplying both sides by 14.5 gets x on its own.
   x = 14.5 × sin 28°
5. Use the calculator (degree mode). First find sin 28°, then finish the calculation.
   sin 28° = 0.46947…
   x = 14.5 × 0.46947… = 6.80733…
6. Round to 1 d.p.. Give the answer to the accuracy asked, with the unit.
   x = 6.8 cm

Check. The other acute angle is 90° − 28° = 62°. From that angle the opposite and adjacent sides swap, so we use cos: x = 14.5 × cos 62° = 6.80733…, 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.

Marks. 1 mark for sin 28° = x ÷ 14.5, 1 mark for the answer 6.8.

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.
   9.2 cm is the opposite.
   13.5 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 = 9.213.5 = 0.68148…
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(9.213.5) = tan⁻¹(0.68148…) = 34.27370…°
5. Round to 1 d.p. Give the angle to the accuracy asked, with the degree sign.
   x = 34.3°

Check. Work forwards: tan 34.3° = 0.68 and the ratio was 0.68. 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.

Marks. 1 mark for tan x = 9.2 ÷ 13.5, 1 mark for using tan⁻¹, 1 mark for the answer 34.3°.

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 ship’s journey makes a right-angled triangle. Use Pythagoras for the distance and tan for the bearing.

1. Distance. The shorter sides are 15 km and 8 km.
   15² + 8² = 225 + 64 = 289, so distance = √289 = 17 km
2. Angle from north. East is 15 and north is 8. At the start the angle between north and the line to the ship has opposite = 15, adjacent = 8.
   tan x = 158 = 1.875
   x = 61.92751…°
3. Bearing. Write three figures, to the nearest degree.
   Bearing = 062°

Check. The ship is further east than north, so the bearing should be between 045° and 090°. 062° fits. ✓

Watch out. East before north: the bearing is from north, so use opposite = east.

Marks. 1 mark for 8² + 15², 1 mark for 17, 1 mark for tan⁻¹(15 ÷ 8), 1 mark for 062°.

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
   tan 30° = √33
2. Put them into the calculation. Replace each trig value by its exact value.
   sin 60° × tan 30° = √32 × √33
3. Work it out exactly. To multiply fractions, multiply the tops and multiply the bottoms. Then simplify.
   √32 × √33 = 12

Check. Decimal check: 0.86602… × 0.57735… = 0.5, and 12 = 0.5. ✓

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.

Marks. 1 mark for each correct exact value, and the product 12.

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

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

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

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

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

Marks. 1 mark for angle C = 61°, 1 mark for the sine rule with correct values, 1 mark for rearranging, 1 mark for the answer 10.8.

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

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

1. Find the angle at B. Draw a north line at B. The back bearing from B to A is 040° + 180° = 220°. The second leg is on 100°.
   Angle ABC = 220° − 100° = 120°
2. Name the sides for the cosine rule. The angle is at B, so a = AC (opposite B), b = AB = 5 km and c = BC = 7 km.
   a = AC, b = 5, c = 7
3. Spot the angle and the two sides around it. The cosine rule is used when we know two sides and the angle between them. The side we want is the one opposite that angle.
   Known: b = 5 km, c = 7 km, angle B = 120° between them.
   Wanted: a, opposite angle B.
4. Write the cosine rule and put the numbers in. Square the wanted side on the left; square the other two sides on the right, then take away 2 × side × side × cos(angle).
   a² = b² + c² − 2bc cos B
   a² = 5² + 7² − 2 × 5 × 7 × cos 120°
5. Work out the two squares and add. This is the first part, b² + c².
   5² + 7² = 25 + 49 = 74
6. Work out the last part on its own. Multiply everything together first. Do this before subtracting (BIDMAS: multiplication comes before subtraction).
   cos 120° = −0.5
   2 × 5 × 7 × −0.5 = −35
7. Subtract. This gives a².
   a² = 74 − −35 = 109
8. Square root, then round to 1 d.p.. We want a, not a².
   a = √109 = 10.44030…
   a = 10.4 km

Check. One side of a triangle is between the difference and the sum of the other two: 2 is less than 10.4, which is less than 12. Angle B is more than 90°, so a² should be more than b² + c²: 109 is more than 74. ✓

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

Marks. 1 mark for angle ABC = 120°, 1 mark for the cosine rule with 2 × 5 × 7 cos 120°, 1 mark for AC² = 109, 1 mark for 10.4 km.

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

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

1. Write the area formula and put the numbers in. The area is ½ × a × b × sin C, and here the angle C is unknown.
   40 = ½ × 9 × 12 × sin C
2. Work out ½ × a × b. This is a plain number, so we can combine it.
   ½ × 9 × 12 = 54
   40 = 54 sin C
3. Divide both sides by 54. This gets sin C on its own.
   sin C = 4054 = 0.74074…
4. Use the inverse sine. The inverse button turns the sine back into an angle.
   C = sin⁻¹(0.74074…) = 47.79455…°
5. Think about a second (obtuse) angle. Sine of an obtuse angle equals sine of 180° minus it, so both give the same area.
   Second angle = 180° − 47.79455…° = 132.20544…°
6. Choose the obtuse angle, round to 1 d.p. The question tells us the angle is obtuse.
   C = 132.2°

Check. Work forwards: ½ × 9 × 12 × sin 132.2° = 40.0 cm², close to 40. ✓

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

Marks. 1 mark for ½ × 9 × 12 sin C = 40, 1 mark for sin C = 0.7407…, 1 mark for 47.8° found, 1 mark for 180° − 47.8° = 132.2°.