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Trigonometry in Right-Angled Triangles

Using sine, cosine and tangent, and the exact trigonometric values, to find sides and angles.

  • 9 key terms
  • All boards

Learning Objectives

  1. 1Label the hypotenuse, opposite and adjacent sides of a right-angled triangle for a given angle.
  2. 2Use sine, cosine and tangent to find a missing side or a missing angle.
  3. 3Recall the exact values of sin, cos and tan for 0, 30, 45, 60 and 90 degrees.
  4. 4Use exact values to solve problems without a calculator, including answers with surds (Higher tier).

Why trigonometry on a non-calculator paper

Pythagoras only links the sides of a right-angled triangle. Trigonometry links the sides to the angles, so it is what you use whenever an angle is involved. A calculator makes the arithmetic easy, so on Paper 1 the questions are designed in other ways: you are given the ratio, such as \(\sin\theta = \dfrac{5}{13}\), or the angle is one of the special angles with exact values, such as \(30^\circ\) or \(45^\circ\). The method is the same as on a calculator paper, so learn it carefully and the other papers will be easy as well.

Finding a side from a ratio

In a right-angled triangle, \(\sin\theta = \dfrac{5}{13}\) and the hypotenuse is 39 cm. Work out the length of the side opposite \(\theta\).

Show the solutionHide the solution
  1. 1 Choose the ratio The question has opposite and hypotenuse, so use \(\sin\theta = \dfrac{O}{H}\).
  2. 2 Substitute \(\dfrac{5}{13} = \dfrac{x}{39}\).
  3. 3 Solve \(x = \dfrac{5}{13} \times 39 = 5 \times 3 = 15\).
  4. 4 Units The opposite side is 15 cm.

Answer15 cm

Finding a missing side

A trigonometry equation is solved by rearranging, and the position of the unknown decides how.

  • Unknown on top

    If the unknown is the numerator, multiply: \(\sin\theta = \dfrac{x}{10}\) gives \(x = 10\sin\theta\).

  • Unknown on the bottom

    If the unknown is the denominator, rearrange and divide: \(\cos\theta = \dfrac{6}{x}\) gives \(x = \dfrac{6}{\cos\theta}\).

  • Write the equation first

    Show \(\sin 30^\circ = \dfrac{x}{8}\) before you rearrange, because this equation earns the first mark.

  • Check

    The hypotenuse is always the longest side, and a side opposite a small angle is short.

Finding a missing angle

To find an angle you use the inverse function, written \(\sin^{-1}\), \(\cos^{-1}\) or \(\tan^{-1}\).

  • The idea

    If \(\sin\theta = \dfrac{1}{2}\), then \(\theta = \sin^{-1}\left(\dfrac{1}{2}\right) = 30^\circ\).

  • Calculator papers

    Enter the ratio and press the inverse key. Papers 2 and 3 expect this.

  • Non-calculator papers

    The ratio will match an exact value, so read it off the table below.

  • Check

    The angle in a right-angled triangle is always less than \(90^\circ\).

Using exact values

A right-angled triangle has a hypotenuse of 8 cm and an angle of \(30^\circ\). Work out the length of the side opposite the \(30^\circ\) angle.

Show the solutionHide the solution
  1. 1 Choose the ratio Opposite and hypotenuse, so \(\sin 30^\circ = \dfrac{x}{8}\).
  2. 2 Use the exact value \(\sin 30^\circ = \dfrac{1}{2}\), so \(\dfrac{1}{2} = \dfrac{x}{8}\).
  3. 3 Solve \(x = \dfrac{1}{2} \times 8 = 4\) cm.
  4. 4 Check The side opposite a \(30^\circ\) angle is half the hypotenuse, which is a useful fact to remember.

Answer4 cm

Finding an angle

A right-angled triangle has opposite side 5 cm and adjacent side 5 cm. Work out the angle \(\theta\).

Show the solutionHide the solution
  1. 1 Choose the ratio Opposite and adjacent, so use \(\tan\theta = \dfrac{O}{A}\).
  2. 2 Substitute \(\tan\theta = \dfrac{5}{5} = 1\).
  3. 3 Use the exact values \(\tan 45^\circ = 1\), so \(\theta = 45^\circ\).
  4. 4 Check Two equal short sides make an isosceles right-angled triangle, with two \(45^\circ\) angles.

Answer\(45^\circ\)

Answers with surds (Higher tier)

When the angle is \(45^\circ\), \(60^\circ\) or \(30^\circ\) and the side is not a multiple that cancels, the answer contains a surd.

  • Tan 60

    \(\tan 60^\circ = \sqrt{3}\), so a triangle with adjacent 5 cm and angle \(60^\circ\) has opposite \(5\sqrt{3}\) cm.

  • Sin 45

    With hypotenuse 10 cm and angle \(45^\circ\), the opposite is \(10 \times \dfrac{\sqrt{2}}{2} = 5\sqrt{2}\) cm.

  • Simplify

    Cancel before you multiply, so \(10 \times \dfrac{\sqrt{2}}{2}\) becomes \(5\sqrt{2}\).

  • Rationalising

    Denominators with a surd are tidied up by multiplying the top and bottom by that surd.

A surd answer

(Higher tier) A right-angled triangle has an angle of \(45^\circ\) and a hypotenuse of 10 cm. Work out the length of the side opposite the \(45^\circ\) angle. Give your answer in the form \(a\sqrt{2}\).

Show the solutionHide the solution
  1. 1 Choose the ratio Opposite and hypotenuse, so \(\sin 45^\circ = \dfrac{x}{10}\).
  2. 2 Use the exact value \(\dfrac{\sqrt{2}}{2} = \dfrac{x}{10}\).
  3. 3 Rearrange \(x = 10 \times \dfrac{\sqrt{2}}{2}\).
  4. 4 Simplify \(x = 5\sqrt{2}\) cm.

Answer\(5\sqrt{2}\) cm

What the OCR exam gives you

OCR prints a formulae sheet with the paper, so some formulae are given to you.

  • Given

    The right-angled triangle ratios, written with the sides called \(a\), \(b\) and \(c\), such as \(\sin A = \dfrac{a}{c}\), \(\cos A = \dfrac{b}{c}\) and \(\tan A = \dfrac{a}{b}\) when \(c\) is the hypotenuse.

  • Translate the letters

    The page uses \(a\) for the side opposite angle \(A\) and \(b\) for the adjacent side, so check which sides you have before using it.

  • Not given

    The exact values of sin, cos and tan for \(0^\circ\), \(30^\circ\), \(45^\circ\), \(60^\circ\) and \(90^\circ\) are not on the page, so learn the table.

Test yourself

  1. 1

    What does SOH stand for?

    Show answerHide answer

    \(\sin\theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}}\).

  2. 2

    What does TOA stand for?

    Show answerHide answer

    \(\tan\theta = \dfrac{\text{Opposite}}{\text{Adjacent}}\).

  3. 3

    What is \(\sin 30^\circ\)?

    Show answerHide answer

    \(\dfrac{1}{2}\).

  4. 4

    What is \(\tan 45^\circ\)?

    Show answerHide answer

    1.

  5. 5

    What is \(\cos 60^\circ\)?

    Show answerHide answer

    \(\dfrac{1}{2}\).

Exam technique: trigonometry

Most lost marks are in the first step, choosing the wrong ratio.

  • Label the triangle

    Write O, A and H on the diagram for the angle you are using.

  • Say which ratio

    Write "\(\sin\theta = \dfrac{O}{H}\)" before you put numbers in.

  • Show the equation

    \(\sin 30^\circ = \dfrac{x}{8}\) scores a mark, even if you solve it incorrectly.

  • Check the answer

    A side opposite a small angle should be short, and the hypotenuse is always longest.

Summary and exam focus

  • Label the hypotenuse, opposite and adjacent sides for the angle you are using.
  • SOH CAH TOA: \(\sin = \dfrac{O}{H}\), \(\cos = \dfrac{A}{H}\) and \(\tan = \dfrac{O}{A}\).
  • Use the inverse functions to find an angle.
  • Learn the exact values for \(0^\circ\), \(30^\circ\), \(45^\circ\), \(60^\circ\) and \(90^\circ\), which are needed on a non-calculator paper.
  • Answers with surds are likely at Higher tier.

Exam focus

A right-angled triangle has a hypotenuse of 12 cm and an angle of \(30^\circ\). Work out the length of the side opposite the \(30^\circ\) angle. (3 marks) (3 marks)

Write \(\sin 30^\circ = \dfrac{x}{12}\), then use \(\sin 30^\circ = \dfrac{1}{2}\) to get \(x = 6\) cm. The first line is a mark, and the exact value is another. Do not use Pythagoras, because only one side is known.

Key terms

The words this lesson expects you to use. Each one is linked from the first place it appears above.

Trigonometry
The study of the links between the sides and angles of triangles.
Sine
The ratio of the opposite side to the hypotenuse in a right-angled triangle.
Cosine
The ratio of the adjacent side to the hypotenuse in a right-angled triangle.
Tangent
The ratio of the opposite side to the adjacent side in a right-angled triangle.
Opposite
The side across from the angle being used.
Adjacent
The side next to the angle being used, other than the hypotenuse.
Inverse function
A function that undoes another, such as sin\(^{-1}\), which turns a ratio back into an angle.
Exact value
A value written as a fraction or surd instead of a rounded decimal.
SOH CAH TOA
A memory aid for the three trigonometric ratios.

Questions and answers

16 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Calculate 4 marks Stretch

The diagram shows a right-angled triangle with an angle of \(45^\circ\). (a) Calculate the value of \(x\). [2 marks] (b) Calculate the length of the hypotenuse. Give your answer in the form \(a\sqrt{2}\). [2 marks]

A right-angled triangle with a 45 degree angle, a base of 6 cm and the opposite side labelled x.

Mark scheme — 4 marks available

  • (a) \(\tan 45^\circ = \dfrac{x}{6}\) with \(\tan 45^\circ = 1\) — M1
  • (a) 6 — A1
  • (b) \(6^2 + 6^2 = 72\) — M1
  • (b) \(6\sqrt{2}\) — A1

Model answer

(a) \(\tan 45^\circ = \dfrac{x}{6}\) and \(\tan 45^\circ = 1\), so \(x = 6\). (b) \(\text{hypotenuse}^2 = 6^2 + 6^2 = 72\), so the hypotenuse is \(\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}\) cm.

2. Exam question Write down 2 marks Easier

(a) Write down the exact value of \(\sin 60^\circ\). [1 mark] (b) Write down the exact value of \(\cos 30^\circ\). [1 mark]

Mark scheme — 2 marks available

  • (a) \(\dfrac{\sqrt{3}}{2}\) — B1
  • (b) \(\dfrac{\sqrt{3}}{2}\) — B1

Model answer

(a) \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\). (b) \(\cos 30^\circ = \dfrac{\sqrt{3}}{2}\).

3. Exam question Calculate 3 marks Core

A right-angled triangle has a hypotenuse of 18 cm and an angle of \(30^\circ\). Calculate the length of the side opposite the \(30^\circ\) angle.

Mark scheme — 3 marks available

  • \(\sin 30^\circ = \dfrac{x}{18}\) — M1
  • \(\dfrac{1}{2} \times 18\) — M1
  • 9 cm — A1

Model answer

\(\sin 30^\circ = \dfrac{x}{18}\) and \(\sin 30^\circ = \dfrac{1}{2}\), so \(x = 9\) cm.

4. Exam question Calculate 3 marks Core

In a right-angled triangle, the side opposite angle \(\theta\) is 4 cm and the hypotenuse is 8 cm. Calculate the size of angle \(\theta\).

Mark scheme — 3 marks available

  • \(\sin\theta = \dfrac{4}{8}\) — M1
  • \(\dfrac{1}{2}\) — M1
  • \(30^\circ\) — A1

Model answer

\(\sin\theta = \dfrac{4}{8} = \dfrac{1}{2}\), so \(\theta = 30^\circ\).

5. Exam question Calculate 3 marks Core

A vertical flagpole is 10 m tall. On level ground it casts a shadow 10 m long. Calculate the angle of elevation of the Sun.

Mark scheme — 3 marks available

  • \(\tan\theta = \dfrac{10}{10}\) — M1
  • \(\tan\theta = 1\) — M1
  • \(45^\circ\) — A1

Model answer

The flagpole and the shadow form a right-angled triangle, so \(\tan\theta = \dfrac{10}{10} = 1\) and \(\theta = 45^\circ\).

6. Exam question Calculate 3 marks Stretch

A right-angled triangle has an angle of \(60^\circ\) and a hypotenuse of 12 cm. Calculate the length of the side opposite the \(60^\circ\) angle. Give your answer in the form \(a\sqrt{3}\).

Mark scheme — 3 marks available

  • \(\sin 60^\circ = \dfrac{x}{12}\) — M1
  • \(12 \times \dfrac{\sqrt{3}}{2}\) — M1
  • \(6\sqrt{3}\) cm — A1

Model answer

\(\sin 60^\circ = \dfrac{x}{12}\), so \(x = 12 \times \dfrac{\sqrt{3}}{2} = 6\sqrt{3}\) cm.

7. Exam question Show that 3 marks Stretch

An equilateral triangle has sides of length 2 cm. By splitting the triangle in half, show that \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\).

Mark scheme — 3 marks available

  • Right-angled triangle with hypotenuse 2 and base 1 — M1
  • Height \(= \sqrt{3}\) — M1
  • \(\sin 60^\circ = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{\sqrt{3}}{2}\) — A1

Model answer

Splitting the triangle gives a right-angled triangle with hypotenuse 2 cm and a base of 1 cm. The height is \(\sqrt{2^2 - 1^2} = \sqrt{3}\) cm. The angle at the top of the original triangle is \(60^\circ\), so the angle opposite the height is \(60^\circ\), and \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\), as required.

8. Multiple choice 1 mark Easier

What is the formula for \(\sin\theta\) in a right-angled triangle?

  1. A \(\dfrac{\text{Adjacent}}{\text{Hypotenuse}}\)
  2. B \(\dfrac{\text{Opposite}}{\text{Hypotenuse}}\) Correct
  3. C \(\dfrac{\text{Opposite}}{\text{Adjacent}}\)
  4. D \(\dfrac{\text{Hypotenuse}}{\text{Opposite}}\)

Why: SOH: sine is opposite over hypotenuse.

9. Multiple choice 1 mark Easier

Which ratio links the opposite side and the adjacent side?

  1. A Tangent Correct
  2. B Sine
  3. C Cosine
  4. D Pythagoras

Why: TOA: tangent is opposite over adjacent.

10. Multiple choice 1 mark Easier

What is the exact value of \(\sin 30^\circ\)?

  1. A \(\dfrac{\sqrt{3}}{2}\)
  2. B 1
  3. C \(\dfrac{\sqrt{2}}{2}\)
  4. D \(\dfrac{1}{2}\) Correct

Why: This is one of the exact values you must learn: \(\sin 30^\circ = \dfrac{1}{2}\).

11. Multiple choice 1 mark Easier

What is the exact value of \(\tan 45^\circ\)?

  1. A 0
  2. B \(\dfrac{1}{2}\)
  3. C 1 Correct
  4. D \(\sqrt{3}\)

Why: At \(45^\circ\) the opposite and adjacent sides are equal, so \(\tan 45^\circ = 1\).

12. Multiple choice 1 mark Core

A right-angled triangle has a hypotenuse of 8 cm and an angle of \(30^\circ\). What is the length of the side opposite the \(30^\circ\) angle?

  1. A 16 cm
  2. B 4 cm Correct
  3. C \(4\sqrt{3}\) cm
  4. D 2 cm

Why: \(\sin 30^\circ = \dfrac{x}{8}\), so \(x = \dfrac{1}{2} \times 8 = 4\).

13. Multiple choice 1 mark Core

A right-angled triangle has opposite side 5 cm and adjacent side 5 cm. What is the angle?

  1. A \(45^\circ\) Correct
  2. B \(30^\circ\)
  3. C \(60^\circ\)
  4. D \(90^\circ\)

Why: \(\tan\theta = \dfrac{5}{5} = 1\), so \(\theta = 45^\circ\).

14. Multiple choice 1 mark Core

In a right-angled triangle, \(\sin\theta = \dfrac{5}{13}\) and the hypotenuse is 39 cm. What is the opposite side?

  1. A 3 cm
  2. B 195 cm
  3. C 5 cm
  4. D 15 cm Correct

Why: \(x = \dfrac{5}{13} \times 39 = 5 \times 3 = 15\).

15. Multiple choice 1 mark Stretch

A right-angled triangle has an angle of \(60^\circ\) and an adjacent side of 5 cm. What is the opposite side? (\(\tan 60^\circ = \sqrt{3}\))

  1. A \(\dfrac{5\sqrt{3}}{3}\) cm
  2. B \(\dfrac{5}{2}\) cm
  3. C \(5\sqrt{3}\) cm Correct
  4. D \(5\sqrt{2}\) cm

Why: \(\tan 60^\circ = \dfrac{x}{5}\), so \(x = 5\sqrt{3}\).

16. Multiple choice 1 mark Stretch

A right-angled triangle has an angle of \(45^\circ\) and a hypotenuse of 10 cm. What is the opposite side?

  1. A \(10\sqrt{2}\) cm
  2. B \(5\sqrt{2}\) cm Correct
  3. C \(5\sqrt{3}\) cm
  4. D 5 cm

Why: \(x = 10 \sin 45^\circ = 10 \times \dfrac{\sqrt{2}}{2} = 5\sqrt{2}\).