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Exam questions · Maths · Further Trigonometry

The Sine Rule

  • 6 exam questions
  • 18 marks
  • 9 quick checks
  1. 1 Calculate [3 marks]

    The diagram is not drawn to scale. Calculate the length of \(AC\). Give your answer as a surd. [3 marks]

    A triangle diagram showing a triangle with two angles and one side.
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    Model answer

    \(\dfrac{x}{\sin 120^\circ} = \dfrac{4}{\sin 30^\circ}\), so \(x = \dfrac{4 \times \frac{\sqrt{3}}{2}}{\frac{1}{2}} = 4\sqrt{3}\) cm.

    Mark scheme

    • Uses \(\dfrac{x}{\sin 120^\circ} = \dfrac{4}{\sin 30^\circ}\) — M1
    • Uses \(\sin 120^\circ = \dfrac{\sqrt{3}}{2}\) and \(\sin 30^\circ = \dfrac{1}{2}\) — M1
    • \(4\sqrt{3}\) — A1
  2. 2 Calculate [3 marks]

    In triangle \(ABC\), angle \(A = 45^\circ\), \(a = 6\) cm and \(b = 3\sqrt{2}\) cm. Angle \(B\) is acute. Calculate the size of angle \(B\). [3 marks]

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    Model answer

    \(\dfrac{\sin B}{3\sqrt{2}} = \dfrac{\sin 45^\circ}{6}\), so \(\sin B = \dfrac{3\sqrt{2} \times \frac{\sqrt{2}}{2}}{6} = \dfrac{3}{6} = \dfrac{1}{2}\). So \(B = 30^\circ\).

    Mark scheme

    • \(\dfrac{\sin B}{3\sqrt{2}} = \dfrac{\sin 45^\circ}{6}\) or equivalent — M1
    • \(\sin B = \dfrac{1}{2}\) — M1
    • \(30\) — A1
  3. 3 Calculate [3 marks]

    In triangle \(ABC\), angle \(A = 120^\circ\), angle \(B = 30^\circ\) and \(a = 6\sqrt{3}\) cm. Calculate the length of \(b\). [3 marks]

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    Model answer

    \(\dfrac{b}{\sin 30^\circ} = \dfrac{6\sqrt{3}}{\sin 120^\circ}\), so \(b = \dfrac{6\sqrt{3} \times \frac{1}{2}}{\frac{\sqrt{3}}{2}} = 6\) cm.

    Mark scheme

    • \(\dfrac{b}{\sin 30^\circ} = \dfrac{6\sqrt{3}}{\sin 120^\circ}\) — M1
    • Uses \(\sin 30^\circ = \dfrac{1}{2}\) and \(\sin 120^\circ = \dfrac{\sqrt{3}}{2}\) — M1
    • \(6\) — A1
  4. 4 Calculate [3 marks]

    The diagram is not drawn to scale. The diagram shows a triangular park \(PQR\). Calculate the length of \(PR\). Give your answer in surd form. [3 marks]

    A triangle diagram showing a triangular park PQR.
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    Model answer

    \(\dfrac{x}{\sin 45^\circ} = \dfrac{5}{\sin 30^\circ}\), so \(x = \dfrac{5 \times \frac{\sqrt{2}}{2}}{\frac{1}{2}} = 5\sqrt{2}\) m.

    Mark scheme

    • \(\dfrac{x}{\sin 45^\circ} = \dfrac{5}{\sin 30^\circ}\) — M1
    • Uses the exact values of \(\sin 45^\circ\) and \(\sin 30^\circ\) — M1
    • \(5\sqrt{2}\) — A1
  5. 5 Explain [3 marks]

    In triangle \(ABC\), angle \(A = 45^\circ\), \(a = 6\) cm and \(b = 6\sqrt{3}\) cm. Show that no such triangle exists. [3 marks]

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    Model answer

    \(\sin B = \dfrac{6\sqrt{3} \times \sin 45^\circ}{6} = \sqrt{3} \times \dfrac{\sqrt{2}}{2} = \dfrac{\sqrt{6}}{2}\). Since \(\sqrt{6} > 2\), \(\sin B > 1\), which is impossible, so the triangle does not exist.

    Mark scheme

    • \(\dfrac{\sin B}{6\sqrt{3}} = \dfrac{\sin 45^\circ}{6}\) — M1
    • \(\sin B = \dfrac{\sqrt{6}}{2}\) — A1
    • States that \(\sqrt{6} > 2\), so \(\sin B > 1\), which is impossible — B1
  6. 6 Show that [3 marks]

    In triangle \(ABC\), angle \(A = 60^\circ\), angle \(B = 45^\circ\) and \(a = 3\sqrt{3}\) cm. Show that \(b = 3\sqrt{2}\) cm. [3 marks]

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    Model answer

    \(\dfrac{b}{\sin 45^\circ} = \dfrac{3\sqrt{3}}{\sin 60^\circ}\), so \(b = \dfrac{3\sqrt{3} \times \frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}} = 3\sqrt{2}\).

    Mark scheme

    • \(\dfrac{b}{\sin 45^\circ} = \dfrac{3\sqrt{3}}{\sin 60^\circ}\) — M1
    • \(b = \dfrac{3\sqrt{3} \times \frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}}\) — M1
    • \(3\sqrt{2}\), with the working shown to the end — B1

Quick check

  1. 1

    In triangle \(ABC\), which side is opposite angle \(A\)?

    1. A\(b\)
    2. B\(c\)
    3. C\(a\)
    4. D\(AB\)
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    C: \(a\)

    Each side is opposite the angle with the same letter.

  2. 2

    Which is the sine rule for finding a side?

    1. A\(a^2 = b^2 + c^2 - 2bc\cos A\)
    2. B\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B}\)
    3. C\(\dfrac{1}{2}ab\sin C\)
    4. D\(a\sin A = b\sin B\)
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    B: \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B}\)

    The sine rule says that side over the sine of its opposite angle is constant.

  3. 3

    What do you need to use the sine rule?

    1. AA side and its opposite angle, plus one more side or angle
    2. BThree sides
    3. CTwo sides and the angle between them
    4. DA right angle
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    A: A side and its opposite angle, plus one more side or angle

    The sine rule needs a matching pair.

  4. 4

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

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

    This is one of the exact values.

  5. 5

    In triangle \(ABC\), \(A = 30^\circ\), \(B = 90^\circ\) and \(a = 5\). What is \(b\)?

    1. A5
    2. B\(\dfrac{5}{2}\)
    3. C10
    4. D\(5\sqrt{3}\)
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    C: 10

    \(\dfrac{b}{\sin 90^\circ} = \dfrac{5}{\sin 30^\circ}\), so \(b = \dfrac{5}{\frac{1}{2}} = 10\).

  6. 6

    In triangle \(ABC\), \(A = 30^\circ\), \(B = 45^\circ\) and \(a = 4\). What is \(b\)?

    1. A\(2\sqrt{2}\)
    2. B\(4\sqrt{2}\)
    3. C\(8\)
    4. D\(4\sqrt{3}\)
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    B: \(4\sqrt{2}\)

    \(b = \dfrac{4\sin 45^\circ}{\sin 30^\circ} = \dfrac{4 \times \frac{\sqrt{2}}{2}}{\frac{1}{2}} = 4\sqrt{2}\).

  7. 7

    In triangle \(ABC\), \(a = 6\), \(b = 6\sqrt{2}\) and \(A = 30^\circ\). What is \(\sin B\)?

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

    \(\sin B = \dfrac{b\sin A}{a} = \dfrac{6\sqrt{2} \times \frac{1}{2}}{6} = \dfrac{\sqrt{2}}{2}\).

  8. 8

    In triangle \(ABC\), \(A = 60^\circ\) and \(B = 45^\circ\). What is angle \(C\)?

    1. A\(105^\circ\)
    2. B\(15^\circ\)
    3. C\(45^\circ\)
    4. D\(75^\circ\)
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    D: \(75^\circ\)

    \(180 - 60 - 45 = 75\).

  9. 9

    In triangle \(ABC\), \(A = 60^\circ\), \(B = 45^\circ\) and \(a = 3\sqrt{3}\). What is \(b\)?

    1. A\(3\sqrt{3}\)
    2. B\(\dfrac{3\sqrt{2}}{2}\)
    3. C\(3\sqrt{2}\)
    4. D\(6\)
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    C: \(3\sqrt{2}\)

    \(b = \dfrac{3\sqrt{3} \times \frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}} = 3\sqrt{2}\).