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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.
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    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) Write down the exact value of \(\cos 60^\circ\). [1 mark] (b) Write down the exact value of \(\sin 90^\circ\). [1 mark]

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    Model answer

    (a) \(\cos 60^\circ = \dfrac{1}{2}\). (b) \(\sin 90^\circ = 1\).

    Mark scheme

    • (a) \(\dfrac{1}{2}\) — B1
    • (b) 1 — B1
  3. 3 Work out [2 marks]

    In a right-angled triangle, \(\sin\theta = \dfrac{3}{5}\) and the hypotenuse is 20 cm. Work out the length of the side opposite angle \(\theta\).

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    Model answer

    \(\dfrac{x}{20} = \dfrac{3}{5}\), so \(x = \dfrac{3}{5} \times 20 = 12\) cm.

    Mark scheme

    • \(\dfrac{3}{5} \times 20\) — M1
    • 12 cm — A1
  4. 4 Work out [3 marks]

    A ramp is 6 m long and makes an angle of \(30^\circ\) with the horizontal ground. Work out the height of the top of the ramp above the ground.

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    Model answer

    The height is opposite the \(30^\circ\) angle and the ramp is the hypotenuse, so \(\sin 30^\circ = \dfrac{h}{6}\). Then \(h = \dfrac{1}{2} \times 6 = 3\) m.

    Mark scheme

    • \(\sin 30^\circ = \dfrac{h}{6}\) — M1
    • \(\dfrac{1}{2} \times 6\) — M1
    • 3 m — A1
  5. 5 Work out [3 marks]

    In a right-angled triangle, the side opposite angle \(\theta\) is 6 cm and the hypotenuse is 12 cm. Work out the size of angle \(\theta\).

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    Model answer

    \(\sin\theta = \dfrac{6}{12} = \dfrac{1}{2}\), so \(\theta = 30^\circ\).

    Mark scheme

    • \(\sin\theta = \dfrac{6}{12}\) — M1
    • \(\dfrac{1}{2}\) — M1
    • \(30^\circ\) — A1
  6. 6 Work out [4 marks]

    A right-angled triangle has an angle of \(60^\circ\) and a hypotenuse of 10 cm. (a) Work out the length of the side adjacent to the \(60^\circ\) angle. [2 marks] (b) Work out the length of the side opposite the \(60^\circ\) angle. Give your answer in the form \(a\sqrt{3}\). [2 marks]

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    Model answer

    (a) \(\cos 60^\circ = \dfrac{x}{10}\), so \(x = \dfrac{1}{2} \times 10 = 5\) cm. (b) \(\sin 60^\circ = \dfrac{y}{10}\), so \(y = \dfrac{\sqrt{3}}{2} \times 10 = 5\sqrt{3}\) cm.

    Mark scheme

    • (a) \(\cos 60^\circ = \dfrac{1}{2}\) used — M1
    • (a) 5 — A1
    • (b) \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\) used — M1
    • (b) \(5\sqrt{3}\) — 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 \(a\sqrt{3}\).

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    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\(\dfrac{\text{Hypotenuse}}{\text{Opposite}}\)
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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

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