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Right-angled trigonometry

Use Pythagoras and sine, cosine and tangent to solve right-triangle problems.

12 free questions · Notes · Worked solutions

Practise this topic

What you will practise

  • Identify the hypotenuse and opposite/adjacent sides.
  • Apply Pythagoras' theorem to find a missing side.
  • Choose and use sin, cos or tan for a right triangle.
  • Find angles and lengths, rounding as requested.

Revision notes

Pythagoras

In a right-angled triangle, a²+b²=c² where c is the hypotenuse, the longest side opposite the right angle. Rearrange by taking a square root after isolating the unknown square.

SOH CAH TOA

Relative to an acute angle θ: sinθ=opposite/hypotenuse, cosθ=adjacent/hypotenuse, tanθ=opposite/adjacent. Identify sides from the chosen angle before selecting a ratio.

Finding angles

Use inverse trigonometric functions: θ=sin⁻¹(ratio), cos⁻¹(ratio) or tan⁻¹(ratio). Ensure the calculator is in degrees when angles are requested in degrees.

Common checks

The hypotenuse must be longer than either shorter side. For an acute angle, sine and cosine are between 0 and 1. Keep full calculator precision until the final rounding.

A worked example

A right triangle has hypotenuse 13 cm and one shorter side 5 cm. Find the other side, then find the angle opposite the 5 cm side to the nearest degree.

  1. Pythagoras gives other side²=13²−5²=169−25=144.
  2. The other side is 12 cm.
  3. For the required angle θ, sinθ=5/13.
  4. θ=sin⁻¹(5/13)≈22.6°, so 23°.

12 cm; 23°

Try it yourself

These are original practice questions, not official past-paper questions. Practice points count correct final answers; working is not automatically assessed.

Choose a question to begin.
Question 1 · Standard1 practice point

In a right triangle, relative to angle θ, which ratio equals opposite/hypotenuse?

Need a hint?

Recall SOH CAH TOA.

Worked solution
  1. SOH means sine=opposite/hypotenuse.
Question 2 · Standard1 practice point

A right triangle has shorter sides 6 cm and 8 cm. What is the hypotenuse?

Need a hint?

Use a²+b²=c².

Worked solution
  1. c²=6²+8²=36+64=100.
  2. c=10 cm.
Question 3 · Standard1 practice point

If cos θ=0.5 and θ is acute, what is θ in degrees?

Need a hint?

Use inverse cosine or a standard exact-angle value.

Worked solution
  1. θ=cos⁻¹(0.5).
  2. For an acute angle, θ=60°.
Question 4 · Standard1 practice point

Which side is the hypotenuse?

Need a hint?

It does not depend on which acute angle is chosen.

Worked solution
  1. The hypotenuse is always opposite the 90° angle.
Question 5 · Foundation2 practice points

A right triangle has legs 9 cm and 12 cm. Find the hypotenuse.

Need a hint?

Square the legs, add, then take the square root.

Worked solution
  1. c²=9²+12²=81+144=225.
  2. c=15 cm.
Question 6 · Standard2 practice points

In a right triangle, the hypotenuse is 10 cm and an acute angle is 30°. Find the side opposite that angle.

Need a hint?

Use sin 30°=opposite/10.

Worked solution
  1. Opposite=10 sin30°.
  2. sin30°=0.5, so opposite=5 cm.
Question 7 · Foundation1 practice point

In a right triangle, the side adjacent to angle θ is 7 cm and the hypotenuse is 10 cm. Find cos θ.

Need a hint?

Cosine is adjacent divided by hypotenuse.

Worked solution
  1. cosθ=7/10.
  2. cosθ=0.7.
Question 8 · Standard2 practice points

A 5 m ladder leans against a vertical wall. Its foot is 3 m from the wall. How high up the wall does it reach?

Need a hint?

The ladder is the hypotenuse.

Worked solution
  1. Height²=5²−3²=25−9=16.
  2. Height=4 m.
Question 9 · Foundation1 practice point

A right triangle has opposite side 6 cm and adjacent side 8 cm relative to θ. Find tan θ.

Need a hint?

Tangent is opposite divided by adjacent.

Worked solution
  1. tanθ=6/8.
  2. tanθ=0.75.
Question 10 · Stretch3 practice points

A right triangle has opposite side 5 cm and hypotenuse 13 cm relative to θ. Find θ to the nearest degree.

Need a hint?

Use sinθ=5/13, then inverse sine in degree mode.

Worked solution
  1. sinθ=5/13≈0.3846.
  2. θ=sin⁻¹(5/13)≈22.6°.
  3. To the nearest degree, θ=23°.
Question 11 · Standard2 practice points

A tree casts a 6 m shadow. The angle of elevation from the shadow tip to its top is 45°. Find the tree's height.

Need a hint?

Opposite and adjacent are equal when tan45°=1.

Worked solution
  1. tan45°=height/6.
  2. 1=height/6, so height=6 m.
Question 12 · Stretch3 practice points

A ramp rises 1.2 m over a horizontal distance of 5 m. Find its angle to the horizontal to the nearest degree.

Need a hint?

Use tangent = vertical rise/horizontal run.

Worked solution
  1. tanθ=1.2/5=0.24.
  2. θ=tan⁻¹(0.24)≈13.5°.
  3. To the nearest degree, θ=13°.

Coverage and sources

Selected IGCSE concepts, with a mix of introductory and more demanding questions. Check your own specification. AI drafted; independent teacher review is pending. Editorial details.

Cambridge IGCSE Mathematics 0580 syllabus

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