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Exam questions · Maths · Geometry and Measures

Trigonometry in Right-Angled Triangles

  • 6 exam questions
  • 20 marks
  • 9 quick checks
  1. 1 Work out [5 marks]

    \(ABC\) 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)

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    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.

    Mark scheme

    • (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
  2. 2 Write down [2 marks]

    (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)

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    Model answer

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

    Mark scheme

    • (a) 0.700 — B1
    • (b) 23.6 — B1
  3. 3 Work out [3 marks]

    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.

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    Model answer

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

    Mark scheme

    • \(\sin 72^\circ = \dfrac{h}{6.5}\) — M1
    • \(6.5 \times \sin 72^\circ\) — M1
    • 6.18 m — A1
  4. 4 Work out [3 marks]

    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.

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    Model answer

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

    Mark scheme

    • \(\tan\theta = \dfrac{7.2}{9.5}\) — M1
    • \(37.158\ldots\) or \(\tan^{-1}\left(\dfrac{7.2}{9.5}\right)\) — M1
    • \(37.2^\circ\) — A1
  5. 5 Work out [3 marks]

    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.

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    Model answer

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

    Mark scheme

    • \(\tan 27^\circ = \dfrac{8.5}{AB}\) — M1
    • \(\dfrac{8.5}{\tan 27^\circ}\) — M1
    • 16.7 cm — A1
  6. 6 Work out [4 marks]

    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)

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    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.

    Mark scheme

    • (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

Quick check

  1. 1

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

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

    SOH: sine is opposite over hypotenuse.

  2. 2

    Which ratio links the opposite side and the adjacent side?

    1. ATangent
    2. BSine
    3. CCosine
    4. DPythagoras
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    A: Tangent

    TOA: tangent is opposite over adjacent.

  3. 3

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

    1. A\(\dfrac{\sqrt{3}}{2}\)
    2. B1
    3. C\(\dfrac{\sqrt{2}}{2}\)
    4. D\(\dfrac{1}{2}\)
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    D: \(\dfrac{1}{2}\)

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

  4. 4

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

    1. A0
    2. B\(\dfrac{1}{2}\)
    3. C1
    4. D\(\sqrt{3}\)
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    C: 1

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

  5. 5

    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. A16 cm
    2. B4 cm
    3. C\(4\sqrt{3}\) cm
    4. D2 cm
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    B: 4 cm

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

  6. 6

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

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

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

  7. 7

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

    1. A3 cm
    2. B195 cm
    3. C5 cm
    4. D15 cm
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    D: 15 cm

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

  8. 8

    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
    4. D\(5\sqrt{2}\) cm
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    C: \(5\sqrt{3}\) cm

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

  9. 9

    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
    3. C\(5\sqrt{3}\) cm
    4. D5 cm
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    B: \(5\sqrt{2}\) cm

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