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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. 4Check that the calculator is in degree mode, and know when an exact value is wanted instead (Higher tier).

Why trigonometry on a 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 a calculator paper the marks are for choosing the right ratio, entering it correctly and giving the accuracy that is asked for. The exact values for \(30^\circ\), \(45^\circ\) and \(60^\circ\) still matter, because some questions ask for an exact answer, and the same method is needed on the non-calculator paper.

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, with the calculator in degree mode.

  • 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

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 when an exact answer is wanted.
  • 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

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

1. Exam question Work out 5 marks Core

\(ABC\) is a right-angled triangle with the right angle at \(B\). \(AC = 14.2\) cm and angle \(BAC = 38^\circ\). Give your answers correct to 3 significant figures. (a) Work out the length of \(BC\). (3 marks) (b) Work out the length of \(AB\). (2 marks)

Mark scheme — 5 marks available

  • (a) \(\sin 38^\circ = \dfrac{BC}{14.2}\) — M1
  • (a) \(14.2 \times \sin 38^\circ\) — M1
  • (a) 8.74 — A1
  • (b) \(14.2 \times \cos 38^\circ\) — M1
  • (b) 11.2 — A1

Model answer

(a) \(\sin 38^\circ = \dfrac{BC}{14.2}\), so \(BC = 14.2 \times \sin 38^\circ = 8.74\) cm. (b) \(AB = 14.2 \times \cos 38^\circ = 11.2\) cm.

2. Exam question Write down 2 marks Easier

(a) Use your calculator to work out \(\tan 35^\circ\). Give your answer correct to 3 decimal places. (1 mark) (b) \(\sin x = 0.4\). Work out the value of \(x\), correct to 1 decimal place. (1 mark)

Mark scheme — 2 marks available

  • (a) 0.700 — B1
  • (b) 23.6 — B1

Model answer

(a) \(\tan 35^\circ = 0.700\). (b) \(x = \sin^{-1}(0.4) = 23.6^\circ\).

3. Exam question Work out 3 marks Core

A ladder of length 6.5 m leans against a vertical wall. The ladder makes an angle of \(72^\circ\) with the horizontal ground. Work out the height of the top of the ladder above the ground. Give your answer correct to 3 significant figures.

Mark scheme — 3 marks available

  • \(\sin 72^\circ = \dfrac{h}{6.5}\) — M1
  • \(6.5 \times \sin 72^\circ\) — M1
  • 6.18 m — A1

Model answer

\(\sin 72^\circ = \dfrac{h}{6.5}\), so \(h = 6.5 \times \sin 72^\circ = 6.18\) m.

4. Exam question Work out 3 marks Core

In a right-angled triangle, the side opposite angle \(\theta\) is 7.2 cm and the side adjacent to angle \(\theta\) is 9.5 cm. Work out the size of angle \(\theta\). Give your answer correct to 1 decimal place.

Mark scheme — 3 marks available

  • \(\tan\theta = \dfrac{7.2}{9.5}\) — M1
  • \(37.158\ldots\) or \(\tan^{-1}\left(\dfrac{7.2}{9.5}\right)\) — M1
  • \(37.2^\circ\) — A1

Model answer

\(\tan\theta = \dfrac{7.2}{9.5}\), so \(\theta = \tan^{-1}\left(\dfrac{7.2}{9.5}\right) = 37.2^\circ\).

5. Exam question Work out 3 marks Core

Triangle \(ABC\) is right-angled at \(B\). \(BC = 8.5\) cm and angle \(BAC = 27^\circ\). Work out the length of \(AB\). Give your answer correct to 3 significant figures.

Mark scheme — 3 marks available

  • \(\tan 27^\circ = \dfrac{8.5}{AB}\) — M1
  • \(\dfrac{8.5}{\tan 27^\circ}\) — M1
  • 16.7 cm — A1

Model answer

\(\tan 27^\circ = \dfrac{8.5}{AB}\), so \(AB = \dfrac{8.5}{\tan 27^\circ} = 16.7\) cm.

6. Exam question Work out 4 marks Stretch

The point \(A\) is on level ground 60 m from the foot of a vertical tower. The angle of elevation of the top of the tower from \(A\) is \(34^\circ\). Give your answers correct to 3 significant figures. (a) Work out the height of the tower. (2 marks) The point \(B\) is on the same level ground, in line with \(A\) and the tower. The angle of elevation of the top of the tower from \(B\) is \(52^\circ\). (b) Work out the distance from \(B\) to the foot of the tower. (2 marks)

Mark scheme — 4 marks available

  • (a) \(60 \times \tan 34^\circ\) — M1
  • (a) 40.5 — A1
  • (b) \(\dfrac{40.47}{\tan 52^\circ}\) or the full-value equivalent — M1
  • (b) 31.6 — A1

Model answer

(a) \(\tan 34^\circ = \dfrac{h}{60}\), so \(h = 60 \times \tan 34^\circ = 40.5\) m. (b) \(\tan 52^\circ = \dfrac{40.47}{d}\), so \(d = \dfrac{40.47}{\tan 52^\circ} = 31.6\) m.

7. 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.

8. 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.

9. 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: \(\sin 30^\circ = \dfrac{1}{2}\).

10. 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 \(\tan 45^\circ = 1\).

11. 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 \(x = \dfrac{1}{2} \times 8 = 4\).

12. 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 \(\theta = 45^\circ\).

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

14. 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 \(x = 5\sqrt{3}\).

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