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Geometry and measuresFoundation and Higher (some Higher only)

Trigonometry GCSE questions with answers

8 trigonometry questions to try, each with the answer and the full working one tap away. Every question here was made and worked out by the Show Working maths engine, not typed in by hand.

Trigonometry uses sin, cos and tan to find a missing side or angle in a right-angled triangle, including angles of elevation and depression. You also need the exact values of sin, cos and tan for angles like 30°, 45° and 60°, and on Higher you use trigonometry in 3D.

Quiz yourself: 8 questions

Worked example 1: exact values of sin, cos and tan

Write down the exact value of sin 0°.

  1. Think of the graphs. The sin graph starts at 0 and reaches its top, 1, at 90°. The cos graph starts at its top, 1, and is 0 at 90°. tan 0° is 0.
  2. So sin 0° = 0.

Answer 0

01 / Quiz yourselfTry it, then tap to check

Trigonometry questions to try

Work each one out on paper first. Then open the answer and compare your working with ours, step by step.

01 / Exact values of sin, cos and tanBoth tiers

Write down the exact value of sin 45°.

Show the answer and working
  1. Use half of a square with sides 1: its angles are 45°, 45° and 90°, the two shorter sides are 1 and the hypotenuse is √2.
  2. sin 45° = opposite ÷ hypotenuse = 1 ÷ √2, which is √2/2 (multiply the top and bottom by √2). 1/√2 is right too.
  3. So sin 45° = √2/2.

Answer √22

02 / Exact values of sin, cos and tanBoth tiers

Write down the exact value of sin 30°.

Show the answer and working
  1. Use half of an equilateral triangle with sides 2: its angles are 30°, 60° and 90°. The side opposite 30° is 1, the side opposite 60° is √3, and the hypotenuse is 2.
  2. sin 30° = opposite ÷ hypotenuse = 1 ÷ 2 = 1/2.
  3. So sin 30° = 1/2.

Answer 12

03 / Find a side with sin, cos or tanBoth tiers

Work out the length of the side marked x. Give your answer to 1 decimal place.

x5 m54°
Not drawn accurately

You can use a calculator for this one.

Show the answer and working
  1. Label the sides from the 54° angle: the hypotenuse is 5 m, and x is the opposite side.
  2. Opposite and hypotenuse use sin, adjacent and hypotenuse use cos, opposite and adjacent use tan. Here: sin θ = opposite ÷ hypotenuse.
  3. sin 54° = x ÷ 5, so x = 5 × sin 54° = 4.0451…
  4. To 1 decimal place: 4.0 m.

Answer 4.0 m

04 / Find a side with sin, cos or tanBoth tiers

Work out the length of the side marked x. Give your answer to 1 decimal place.

8 cmx63°
Not drawn accurately

You can use a calculator for this one.

Show the answer and working
  1. Label the sides from the 63° angle: the opposite side is 8 cm, and x is the hypotenuse.
  2. Opposite and hypotenuse use sin, adjacent and hypotenuse use cos, opposite and adjacent use tan. Here: sin θ = opposite ÷ hypotenuse.
  3. sin 63° = 8 ÷ x, so x = 8 ÷ sin 63° = 8.9786…
  4. To 1 decimal place: 9.0 cm.

Answer 9.0 cm

05 / Find an angle with sin, cos or tanBoth tiers

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

7 m14 mx
Not drawn accurately

You can use a calculator for this one.

Show the answer and working
  1. From angle x, the sides you know are the adjacent (7 m) and the hypotenuse (14 m).
  2. Those two sides go with cos: cos x = 7 ÷ 14.
  3. Use the inverse on your calculator: x = cos⁻¹(7 ÷ 14) = 60.0000…°
  4. To 1 decimal place: 60.0°.

Answer 60.0°

06 / Find an angle with sin, cos or tanBoth tiers

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

8 cm10 cmx
Not drawn accurately

You can use a calculator for this one.

Show the answer and working
  1. From angle x, the sides you know are the adjacent (8 cm) and the hypotenuse (10 cm).
  2. Those two sides go with cos: cos x = 8 ÷ 10.
  3. Use the inverse on your calculator: x = cos⁻¹(8 ÷ 10) = 36.8699…°
  4. To 1 decimal place: 36.9°.

Answer 36.9°

07 / Angles of elevation and depressionBoth tiers

From a point on level ground 55 m from the foot of a church spire, the angle of elevation of the top is 50°. Work out the height of the church spire. Give your answer to 1 decimal place.

50°55 m
Not drawn accurately

You can use a calculator for this one.

Show the answer and working
  1. The angle of elevation is measured up from the horizontal. tan = opposite ÷ adjacent. From the angle, the height is opposite and the distance is adjacent.
  2. Height = 55 × tan 50° = 65.5464…
  3. To 1 decimal place: 65.5 m.

Answer 65.5 m

08 / Trigonometry in 3DHigher only

The diagram shows a cuboid ABCDEFGH. AB = 15 cm, BC = 7 cm and AE = 10 cm. Work out the size of the angle between the diagonal AG and the base ABCD. Give your answer to 1 decimal place.

15 cm7 cm10 cmABCDEFGH
Not drawn accurately

You can use a calculator for this one.

Show the answer and working
  1. G is directly above C, so the angle between AG and the base is angle GAC, in triangle ACG. It has a right angle at C. Call the angle x.
  2. AC is the diagonal of the base: AC² = 15² + 7² = 274, so AC = 16.5529…
  3. CG = AE = 10 cm. tan x = CG ÷ AC = 10 ÷ 16.5529…
  4. x = tan⁻¹(10 ÷ 16.5529…) = 31.1371…°
  5. To 1 decimal place: 31.1°.

Answer 31.1°

02 / What's covered5 kinds of question

Trigonometry at GCSE and where marks go

  • Exact values of sin, cos and tanBoth tiers
  • Find a side with sin, cos or tanBoth tiers
  • Find an angle with sin, cos or tanBoth tiers
  • Angles of elevation and depressionBoth tiers
  • Trigonometry in 3DHigher only

Worked example 2: trigonometry in 3d

VABCD is a pyramid with a square base ABCD of side 4 cm. M is the centre of the base and V is directly above M, with VM = 20 cm. Work out the size of the angle between the edge VB and the base. Give your answer to 1 decimal place.

4 cm20 cmxABCDVM
Not drawn accurately

You can use a calculator for this one.

  1. V is directly above M, so the angle between VB and the base is angle VBM, in the right-angled triangle VBM.
  2. BD is the diagonal of the base: BD² = 4² + 4² = 32, so BD = 5.6569… and BM, half of it, is 2.8284…
  3. tan x = VM ÷ BM = 20 ÷ 2.8284…
  4. x = tan⁻¹(20 ÷ 2.8284…) = 81.9505…°
  5. To 1 decimal place: 82.0°.

Answer 82.0°

A common mistake

Picking sin, cos or tan before labelling the sides. Label opposite, adjacent and hypotenuse from the angle first. With the opposite 5 cm and the hypotenuse 10 cm you need sin: sin x = 5/10, so x = 30°.

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