OpenRevise

Exam questions · Maths · Geometry and Measures

Trigonometry in Right-Angled Triangles

  • 7 exam questions
  • 21 marks
  • 9 quick checks
  1. 1 Work out [4 marks]

    The diagram shows a right-angled triangle. (a) Work out the value of \(x\). [3 marks] (b) Write down the size of the third angle of the triangle. [1 mark]

    A right-angled triangle with a 30 degree angle, a hypotenuse of 8 cm and the opposite side labelled x.
    Show answerHide answer

    Model answer

    (a) \(\sin 30^\circ = \dfrac{x}{8}\) and \(\sin 30^\circ = \dfrac{1}{2}\), so \(x = 4\) cm. (b) \(180 - 90 - 30 = 60^\circ\).

    Mark scheme

    • (a) \(\sin 30^\circ = \dfrac{x}{8}\) — M1
    • (a) \(\dfrac{1}{2} = \dfrac{x}{8}\) — M1
    • (a) 4 — A1
    • (b) \(60^\circ\) — B1
  2. 2 Write down [2 marks]

    (a) Use your calculator to work out \(\cos 48^\circ\). Give your answer correct to 3 significant figures. [1 mark] (b) \(\tan x = 1.4\). Work out the value of \(x\), correct to 1 decimal place. [1 mark]

    Show answerHide answer

    Model answer

    (a) \(\cos 48^\circ = 0.669\). (b) \(x = \tan^{-1}(1.4) = 54.5^\circ\).

    Mark scheme

    • (a) 0.669 — B1
    • (b) 54.5 — B1
  3. 3 Work out [2 marks]

    In a right-angled triangle, the hypotenuse is 12.4 cm and one of the other angles is \(41^\circ\). Work out the length of the side adjacent to the \(41^\circ\) angle. Give your answer correct to 3 significant figures.

    Show answerHide answer

    Model answer

    \(\cos 41^\circ = \dfrac{x}{12.4}\), so \(x = 12.4 \times \cos 41^\circ = 9.36\) cm.

    Mark scheme

    • \(12.4 \times \cos 41^\circ\) — M1
    • 9.36 cm — A1
  4. 4 Work out [3 marks]

    A ramp is 6 m long and makes an angle of \(14^\circ\) with the horizontal ground. Work out the height of the top of the ramp above the ground. Give your answer correct to 3 significant figures.

    Show answerHide answer

    Model answer

    The height is opposite the \(14^\circ\) angle and the ramp is the hypotenuse, so \(\sin 14^\circ = \dfrac{h}{6}\). Then \(h = 6 \times \sin 14^\circ = 1.45\) m.

    Mark scheme

    • \(\sin 14^\circ = \dfrac{h}{6}\) — M1
    • \(6 \times \sin 14^\circ\) — M1
    • 1.45 m — A1
  5. 5 Work out [3 marks]

    In a right-angled triangle, the side opposite angle \(\theta\) is 6.4 cm and the hypotenuse is 11.5 cm. Work out the size of angle \(\theta\). Give your answer correct to 1 decimal place.

    Show answerHide answer

    Model answer

    \(\sin\theta = \dfrac{6.4}{11.5}\), so \(\theta = \sin^{-1}\left(\dfrac{6.4}{11.5}\right) = 33.8^\circ\).

    Mark scheme

    • \(\sin\theta = \dfrac{6.4}{11.5}\) — M1
    • \(33.816\ldots\) or \(\sin^{-1}\left(\dfrac{6.4}{11.5}\right)\) — M1
    • \(33.8^\circ\) — A1
  6. 6 Work out [4 marks]

    A plane flies 120 km from \(A\) to \(B\) on a bearing of \(065^\circ\). Give your answers correct to 3 significant figures. (a) How far north of \(A\) is \(B\)? [2 marks] (b) How far east of \(A\) is \(B\)? [2 marks]

    Show answerHide answer

    Model answer

    (a) The northward distance is \(120 \times \cos 65^\circ = 50.7\) km. (b) The eastward distance is \(120 \times \sin 65^\circ = 109\) km.

    Mark scheme

    • (a) \(120 \times \cos 65^\circ\) — M1
    • (a) 50.7 — A1
    • (b) \(120 \times \sin 65^\circ\) — M1
    • (b) 109 — A1
  7. 7 Work out [3 marks]

    A right-angled triangle has an angle of \(30^\circ\) and an adjacent side of 6 cm. \(\tan 30^\circ = \dfrac{\sqrt{3}}{3}\). Work out the length of the side opposite the \(30^\circ\) angle. Give your answer in the form \(a\sqrt{3}\).

    Show answerHide answer

    Model answer

    \(\tan 30^\circ = \dfrac{x}{6}\), so \(x = 6 \times \dfrac{\sqrt{3}}{3} = 2\sqrt{3}\) cm.

    Mark scheme

    • \(\tan 30^\circ = \dfrac{x}{6}\) — M1
    • \(6 \times \dfrac{\sqrt{3}}{3}\) — M1
    • \(2\sqrt{3}\) cm — 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\(\dfrac{\text{Hypotenuse}}{\text{Opposite}}\)
    Show answerHide answer

    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
    Show answerHide answer

    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\(\dfrac{1}{2}\)
    Show answerHide answer

    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\(\sqrt{3}\)
    Show answerHide answer

    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
    Show answerHide answer

    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\(90^\circ\)
    Show answerHide answer

    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
    Show answerHide answer

    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
    Show answerHide answer

    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
    Show answerHide answer

    B: \(5\sqrt{2}\) cm

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