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Exam questions · Maths

Further Trigonometry

  • 30 exam questions
  • 98 marks
  • 45 quick checks

Trigonometric Graphs and Exact Values

Just this lesson
  1. 1 Solve [2 marks]

    The diagram shows the graph of \(y = \cos x\) for \(0^\circ \le x \le 360^\circ\), and the line \(y = 0.5\). Solve \(\cos x = 0.5\) for \(0^\circ \le x \le 360^\circ\). (2 marks)

    The graph of cosine from 0 to 360 degrees with the line y equals 0.5 crossing the curve twice.
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    Model answer

    \(\cos 60^\circ = 0.5\), so \(x = 60^\circ\). The cosine graph is symmetrical about \(180^\circ\), so the other solution is \(360 - 60 = 300^\circ\).

    Mark scheme

    • \(60\) — M1
    • \(300\) — A1
  2. 2 Write down [3 marks]

    The diagram shows a graph for \(0^\circ \le x \le 360^\circ\). (a) Write down the equation of the graph. (1 mark) (b) Write down the coordinates of the maximum point. (1 mark) (c) Write down the values of \(x\) where the graph crosses the \(x\)-axis. (1 mark)

    A wave graph between 1 and minus 1 starting at the origin, with a maximum at 90 degrees.
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    Model answer

    (a) \(y = \sin x\). (b) \((90, 1)\). (c) \(x = 0^\circ\), \(180^\circ\) and \(360^\circ\).

    Mark scheme

    • (a) \(y = \sin x\) — B1
    • (b) \((90, 1)\) — B1
    • (c) \(0, 180, 360\) — B1
  3. 3 Write down [3 marks]

    Write down the exact value of (a) \(\sin 150^\circ\) (1 mark) (b) \(\cos 120^\circ\) (1 mark) (c) \(\tan 135^\circ\) (1 mark)

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    Model answer

    (a) \(\dfrac{1}{2}\). (b) \(-\dfrac{1}{2}\). (c) \(-1\).

    Mark scheme

    • (a) \(\dfrac{1}{2}\) — B1
    • (b) \(-\dfrac{1}{2}\) — B1
    • (c) \(-1\) — B1
  4. 4 Solve [3 marks]

    Solve \(2\sin x + \sqrt{3} = 0\) for \(0^\circ \le x \le 360^\circ\). (3 marks)

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    Model answer

    \(\sin x = -\dfrac{\sqrt{3}}{2}\). Sine is negative between \(180^\circ\) and \(360^\circ\), and \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\), so \(x = 180 + 60 = 240^\circ\) or \(x = 360 - 60 = 300^\circ\).

    Mark scheme

    • \(\sin x = -\dfrac{\sqrt{3}}{2}\) — M1
    • \(240\) — A1
    • \(300\) — A1
  5. 5 Solve [3 marks]

    Solve \(\tan x = -1\) for \(0^\circ \le x \le 360^\circ\). (3 marks)

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    Model answer

    \(\tan 45^\circ = 1\), and tangent is negative between \(90^\circ\) and \(180^\circ\), and between \(270^\circ\) and \(360^\circ\). So \(x = 180 - 45 = 135^\circ\) or \(x = 360 - 45 = 315^\circ\).

    Mark scheme

    • Uses \(45^\circ\) as the related angle — M1
    • \(135\) — A1
    • \(315\) — A1
  6. 6 Explain [2 marks]

    (a) Explain why the equation \(\sin x = \dfrac{3}{2}\) has no solutions. (1 mark) (b) How many solutions does \(\cos x = -1\) have for \(0^\circ \le x \le 360^\circ\)? (1 mark)

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    Model answer

    (a) The sine of an angle is never greater than 1. (b) One solution, \(x = 180^\circ\).

    Mark scheme

    • (a) The sine of an angle is never more than 1, or the graph does not go above 1 — C1
    • (b) 1, which is \(x = 180^\circ\) — B1

Quick check

  1. 1

    What is the maximum value of \(y = \sin x\)?

    1. A0
    2. B1
    3. C90
    4. D\(\infty\)
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    B: 1

    The sine graph is a wave between \(-1\) and \(1\).

  2. 2

    What is \(\cos 0^\circ\)?

    1. A1
    2. B0
    3. C\(-1\)
    4. D\(\dfrac{1}{2}\)
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    A: 1

    The cosine graph starts at its maximum, 1.

  3. 3

    Where are the asymptotes of \(y = \tan x\) between \(0^\circ\) and \(360^\circ\)?

    1. A\(x = 0^\circ\) and \(x = 180^\circ\)
    2. B\(x = 45^\circ\) and \(x = 135^\circ\)
    3. C\(x = 180^\circ\) and \(x = 360^\circ\)
    4. D\(x = 90^\circ\) and \(x = 270^\circ\)
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    D: \(x = 90^\circ\) and \(x = 270^\circ\)

    The tangent is undefined at \(90^\circ\) and \(270^\circ\).

  4. 4

    How many solutions does \(\sin x = \dfrac{1}{2}\) have for \(0^\circ \le x \le 360^\circ\)?

    1. A1
    2. B3
    3. C2
    4. D4
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    C: 2

    The line \(y = \dfrac{1}{2}\) crosses the sine curve twice, at \(30^\circ\) and \(150^\circ\).

  5. 5

    If \(\sin 30^\circ = \dfrac{1}{2}\), what is the other solution of \(\sin x = \dfrac{1}{2}\) between \(0^\circ\) and \(360^\circ\)?

    1. A\(120^\circ\)
    2. B\(150^\circ\)
    3. C\(210^\circ\)
    4. D\(330^\circ\)
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    B: \(150^\circ\)

    The second solution is \(180^\circ - 30^\circ = 150^\circ\).

  6. 6

    What is the period of \(y = \tan x\)?

    1. A\(180^\circ\)
    2. B\(360^\circ\)
    3. C\(90^\circ\)
    4. D\(45^\circ\)
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    A: \(180^\circ\)

    The tangent graph repeats every \(180^\circ\).

  7. 7

    Solve \(\cos x = \dfrac{1}{2}\) for \(0^\circ \le x \le 360^\circ\).

    1. A\(60^\circ\) and \(120^\circ\)
    2. B\(30^\circ\) and \(330^\circ\)
    3. C\(60^\circ\) and \(240^\circ\)
    4. D\(60^\circ\) and \(300^\circ\)
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    D: \(60^\circ\) and \(300^\circ\)

    \(\cos 60^\circ = \dfrac{1}{2}\), and the second solution is \(360^\circ - 60^\circ = 300^\circ\).

  8. 8

    Solve \(\sin x = -\dfrac{1}{2}\) for \(0^\circ \le x \le 360^\circ\).

    1. A\(30^\circ\) and \(150^\circ\)
    2. B\(120^\circ\) and \(240^\circ\)
    3. C\(210^\circ\) and \(330^\circ\)
    4. D\(150^\circ\) and \(210^\circ\)
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    C: \(210^\circ\) and \(330^\circ\)

    Sine is negative between \(180^\circ\) and \(360^\circ\), so the solutions are \(180 + 30 = 210\) and \(360 - 30 = 330\).

  9. 9

    Which equation has no solutions?

    1. A\(\cos x = -1\)
    2. B\(\sin x = 1.5\)
    3. C\(\tan x = 5\)
    4. D\(\sin x = 0\)
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    B: \(\sin x = 1.5\)

    Sine and cosine are never greater than 1, so \(\sin x = 1.5\) has no solution.

The Sine Rule

Just this lesson
  1. 1 Work out [3 marks]

    Diagram NOT accurately drawn. Work out the length of \(AC\). Give your answer in the form \(k\sqrt{2}\). (3 marks)

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

    \(\dfrac{x}{\sin 45^\circ} = \dfrac{8}{\sin 30^\circ}\), so \(x = \dfrac{8 \times \frac{\sqrt{2}}{2}}{\frac{1}{2}} = 8\sqrt{2}\) cm.

    Mark scheme

    • \(\dfrac{x}{\sin 45^\circ} = \dfrac{8}{\sin 30^\circ}\) or equivalent — M1
    • Uses \(\sin 45^\circ = \dfrac{\sqrt{2}}{2}\) and \(\sin 30^\circ = \dfrac{1}{2}\) — M1
    • \(8\sqrt{2}\) — A1
  2. 2 Work out [3 marks]

    In triangle \(ABC\), angle \(A = 45^\circ\), \(a = 10\) cm and \(b = 5\sqrt{2}\) cm. Angle \(B\) is acute. Work out the size of angle \(B\). (3 marks)

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    Model answer

    \(\dfrac{\sin B}{5\sqrt{2}} = \dfrac{\sin 45^\circ}{10}\), so \(\sin B = \dfrac{5\sqrt{2} \times \frac{\sqrt{2}}{2}}{10} = \dfrac{5}{10} = \dfrac{1}{2}\). So \(B = 30^\circ\).

    Mark scheme

    • \(\dfrac{\sin B}{5\sqrt{2}} = \dfrac{\sin 45^\circ}{10}\) or equivalent — M1
    • \(\sin B = \dfrac{1}{2}\) — M1
    • \(30\) — A1
  3. 3 Work out [3 marks]

    In triangle \(ABC\), angle \(A = 30^\circ\), angle \(B = 30^\circ\) and \(a = 7\) cm. Work out the length of \(c\). Give your answer in surd form. (3 marks)

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    Model answer

    Angle \(C = 180 - 30 - 30 = 120^\circ\). \(\dfrac{c}{\sin 120^\circ} = \dfrac{7}{\sin 30^\circ}\), so \(c = \dfrac{7 \times \frac{\sqrt{3}}{2}}{\frac{1}{2}} = 7\sqrt{3}\) cm.

    Mark scheme

    • Angle \(C = 120^\circ\) — M1
    • \(\dfrac{c}{\sin 120^\circ} = \dfrac{7}{\sin 30^\circ}\) — M1
    • \(7\sqrt{3}\) — A1
  4. 4 Work out [3 marks]

    Diagram NOT accurately drawn. The diagram shows a triangular park \(PQR\). Work out the length of \(PR\). Give your answer in the form \(k\sqrt{2}\). (3 marks)

    A triangle diagram showing a triangular park PQR.
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    Model answer

    \(\dfrac{x}{\sin 30^\circ} = \dfrac{80}{\sin 45^\circ}\), so \(x = \dfrac{80 \times \frac{1}{2}}{\frac{\sqrt{2}}{2}} = \dfrac{80}{\sqrt{2}} = 40\sqrt{2}\) m.

    Mark scheme

    • \(\dfrac{x}{\sin 30^\circ} = \dfrac{80}{\sin 45^\circ}\) — M1
    • \(x = \dfrac{80}{\sqrt{2}}\) or \(\dfrac{40}{\frac{\sqrt{2}}{2}}\) — M1
    • \(40\sqrt{2}\) — A1
  5. 5 Work out [3 marks]

    In triangle \(ABC\), angle \(A = 30^\circ\), \(a = 5\) cm and \(b = 5\sqrt{3}\) cm. There are two possible sizes of angle \(B\). Work out both of them. (3 marks)

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    Model answer

    \(\sin B = \dfrac{5\sqrt{3} \times \frac{1}{2}}{5} = \dfrac{\sqrt{3}}{2}\). So \(B = 60^\circ\) or \(B = 180 - 60 = 120^\circ\).

    Mark scheme

    • \(\sin B = \dfrac{\sqrt{3}}{2}\) — M1
    • \(60\) — A1
    • \(120\) — A1
  6. 6 Show that [3 marks]

    In triangle \(ABC\), angle \(A = 45^\circ\), angle \(B = 60^\circ\) and \(a = \sqrt{6}\) cm. Show that \(b = 3\) cm. (3 marks)

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    Model answer

    \(\dfrac{b}{\sin 60^\circ} = \dfrac{\sqrt{6}}{\sin 45^\circ}\), so \(b = \dfrac{\sqrt{6} \times \frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}} = \dfrac{\sqrt{18}}{\sqrt{2}} = \sqrt{9} = 3\).

    Mark scheme

    • \(\dfrac{b}{\sin 60^\circ} = \dfrac{\sqrt{6}}{\sin 45^\circ}\) — M1
    • \(b = \dfrac{\sqrt{6} \times \frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}}\) — M1
    • \(\dfrac{\sqrt{18}}{\sqrt{2}} = 3\), with the working shown to the end — C1

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

    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\(\sqrt{2}\)
    Show answerHide answer

    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\(75^\circ\)
    Show answerHide answer

    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\(6\)
    Show answerHide answer

    C: \(3\sqrt{2}\)

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

The Cosine Rule

Just this lesson
  1. 1 Work out [3 marks]

    Diagram NOT accurately drawn. Work out the length of \(BC\). (3 marks)

    A triangle diagram showing a triangle with two sides and the angle between them.
    Show answerHide answer

    Model answer

    \(x^2 = 8^2 + 15^2 - 2 \times 8 \times 15 \times \cos 60^\circ = 64 + 225 - 120 = 169\), so \(x = 13\) cm.

    Mark scheme

    • \(x^2 = 8^2 + 15^2 - 2 \times 8 \times 15 \times \cos 60^\circ\) — M1
    • \(64 + 225 - 120 = 169\) — M1
    • \(13\) — A1
  2. 2 Work out [3 marks]

    A triangle has sides of length 5 cm, 7 cm and 8 cm. Work out the size of the angle opposite the side of length 7 cm. (3 marks)

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    Model answer

    \(\cos A = \dfrac{5^2 + 8^2 - 7^2}{2 \times 5 \times 8} = \dfrac{40}{80} = \dfrac{1}{2}\), so \(A = 60^\circ\).

    Mark scheme

    • \(\cos A = \dfrac{5^2 + 8^2 - 7^2}{2 \times 5 \times 8}\) — M1
    • \(\dfrac{1}{2}\) — A1
    • \(60\) — A1
  3. 3 Work out [3 marks]

    Diagram NOT accurately drawn. Two ships leave a port \(P\) at the same time. Ship \(Q\) sails 50 km and ship \(R\) sails 80 km. The angle between their paths is \(60^\circ\). Work out the distance \(QR\) between the ships. (3 marks)

    A triangle diagram showing two ships leaving a port P.
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    Model answer

    \(x^2 = 50^2 + 80^2 - 2 \times 50 \times 80 \times \cos 60^\circ = 2500 + 6400 - 4000 = 4900\), so \(x = 70\) km.

    Mark scheme

    • \(x^2 = 50^2 + 80^2 - 2 \times 50 \times 80 \times \cos 60^\circ\) — M1
    • \(2500 + 6400 - 4000 = 4900\) — M1
    • \(70\) — A1
  4. 4 Work out [3 marks]

    In triangle \(ABC\), \(AB = AC = 6\) cm and angle \(A = 120^\circ\). Work out the length of \(BC\). Give your answer as a surd in its simplest form. (3 marks)

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    Model answer

    \(BC^2 = 6^2 + 6^2 - 2 \times 6 \times 6 \times \cos 120^\circ = 36 + 36 + 36 = 108\), so \(BC = \sqrt{108} = 6\sqrt{3}\) cm.

    Mark scheme

    • \(BC^2 = 6^2 + 6^2 - 2 \times 6 \times 6 \times \cos 120^\circ\) — M1
    • \(108\) — M1
    • \(6\sqrt{3}\) — A1
  5. 5 Work out [5 marks]

    In triangle \(ABC\), \(AB = 5\) cm, \(BC = 8\) cm and \(AC = 7\) cm. (a) Show that angle \(ABC = 60^\circ\). (3 marks) (b) Work out the exact area of triangle \(ABC\). (2 marks)

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    Model answer

    (a) \(\cos B = \dfrac{5^2 + 8^2 - 7^2}{2 \times 5 \times 8} = \dfrac{40}{80} = \dfrac{1}{2}\), so \(B = 60^\circ\). (b) Area \(= \dfrac{1}{2} \times 5 \times 8 \times \sin 60^\circ = 20 \times \dfrac{\sqrt{3}}{2} = 10\sqrt{3}\) cm\(^2\).

    Mark scheme

    • (a) \(\cos B = \dfrac{5^2 + 8^2 - 7^2}{2 \times 5 \times 8}\) — M1
    • (a) \(\dfrac{40}{80} = \dfrac{1}{2}\) — M1
    • (a) \(B = 60^\circ\), with the conclusion stated — C1
    • (b) \(\dfrac{1}{2} \times 5 \times 8 \times \sin 60^\circ\) — M1
    • (b) \(10\sqrt{3}\) — A1
  6. 6 Work out [3 marks]

    A triangle has sides of length 6 cm, 6 cm and \(6\sqrt{3}\) cm. Work out the size of the largest angle. (3 marks)

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    Model answer

    The largest angle is opposite the longest side. \(\cos A = \dfrac{6^2 + 6^2 - (6\sqrt{3})^2}{2 \times 6 \times 6} = \dfrac{36 + 36 - 108}{72} = -\dfrac{1}{2}\), so \(A = 120^\circ\).

    Mark scheme

    • \(\cos A = \dfrac{6^2 + 6^2 - (6\sqrt{3})^2}{2 \times 6 \times 6}\) — M1
    • \(-\dfrac{1}{2}\) — A1
    • \(120\) — A1

Quick check

  1. 1

    Which is the cosine rule for the side \(a\)?

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

    D: \(a^2 = b^2 + c^2 - 2bc\cos A\)

    The cosine rule takes \(2bc\cos A\) away from the sum of the squares.

  2. 2

    When do you use the cosine rule to find a side?

    1. AWhen you know two angles and a side
    2. BWhen you know a side and its opposite angle
    3. CWhen you know two sides and the angle between them
    4. DWhen the triangle has a right angle
    Show answerHide answer

    C: When you know two sides and the angle between them

    Two sides and the included angle is the cosine rule situation.

  3. 3

    What is \(\cos 60^\circ\)?

    1. A\(\dfrac{\sqrt{3}}{2}\)
    2. B\(\dfrac{1}{2}\)
    3. C\(\dfrac{\sqrt{2}}{2}\)
    4. D1
    Show answerHide answer

    B: \(\dfrac{1}{2}\)

    This is an exact value.

  4. 4

    What does the cosine rule become when \(A = 90^\circ\)?

    1. APythagoras' theorem
    2. BThe sine rule
    3. CThe area formula
    4. D\(a = b + c\)
    Show answerHide answer

    A: Pythagoras' theorem

    \(\cos 90^\circ = 0\), so \(a^2 = b^2 + c^2\).

  5. 5

    In triangle \(ABC\), \(b = 3\), \(c = 8\) and \(A = 60^\circ\). What is \(a^2\)?

    1. A73
    2. B97
    3. C25
    4. D49
    Show answerHide answer

    D: 49

    \(a^2 = 9 + 64 - 2 \times 3 \times 8 \times \frac{1}{2} = 73 - 24 = 49\).

  6. 6

    In triangle \(ABC\), \(b = 6\), \(c = 10\) and \(A = 120^\circ\). What is \(a\)?

    1. A8
    2. B\(\sqrt{76}\)
    3. C14
    4. D16
    Show answerHide answer

    C: 14

    \(a^2 = 36 + 100 - 120 \times (-\frac{1}{2}) = 136 + 60 = 196\), so \(a = 14\).

  7. 7

    A triangle has sides 5, 7 and 8. What is \(\cos\) of the angle opposite the side of length 7?

    1. A\(-\dfrac{1}{2}\)
    2. B\(\dfrac{1}{2}\)
    3. C\(\dfrac{1}{7}\)
    4. D\(\dfrac{3}{5}\)
    Show answerHide answer

    B: \(\dfrac{1}{2}\)

    \(\cos A = \dfrac{5^2 + 8^2 - 7^2}{2 \times 5 \times 8} = \dfrac{40}{80} = \dfrac{1}{2}\).

  8. 8

    A triangle has sides 3, 5 and 7. What is the largest angle?

    1. A\(120^\circ\)
    2. B\(60^\circ\)
    3. C\(150^\circ\)
    4. D\(90^\circ\)
    Show answerHide answer

    A: \(120^\circ\)

    \(\cos A = \dfrac{9 + 25 - 49}{30} = -\dfrac{1}{2}\), so \(A = 120^\circ\).

  9. 9

    A triangle has sides 7, 8 and 13. What is the angle opposite the longest side?

    1. A\(60^\circ\)
    2. B\(90^\circ\)
    3. C\(150^\circ\)
    4. D\(120^\circ\)
    Show answerHide answer

    D: \(120^\circ\)

    \(\cos A = \dfrac{49 + 64 - 169}{112} = \dfrac{-56}{112} = -\dfrac{1}{2}\), so \(A = 120^\circ\).

Area of a Triangle and Segments

Just this lesson
  1. 1 Work out [2 marks]

    Diagram NOT accurately drawn. Work out the area of triangle \(ABC\). (2 marks)

    A triangle diagram showing a triangle with two sides and the angle between them.
    Show answerHide answer

    Model answer

    Area \(= \dfrac{1}{2} \times 8 \times 5 \times \sin 30^\circ = 20 \times \dfrac{1}{2} = 10\) cm\(^2\).

    Mark scheme

    • \(\dfrac{1}{2} \times 8 \times 5 \times \sin 30^\circ\) — M1
    • \(10\) — A1
  2. 2 Work out [3 marks]

    A triangle has two sides of length 12 cm and 5 cm. Its area is 15 cm\(^2\). The angle between the two sides is acute. Work out the size of this angle. (3 marks)

    Show answerHide answer

    Model answer

    \(15 = \dfrac{1}{2} \times 12 \times 5 \times \sin C = 30\sin C\), so \(\sin C = \dfrac{1}{2}\) and \(C = 30^\circ\).

    Mark scheme

    • \(15 = \dfrac{1}{2} \times 12 \times 5 \times \sin C\) — M1
    • \(\sin C = \dfrac{1}{2}\) — M1
    • \(30\) — A1
  3. 3 Work out [3 marks]

    In triangle \(ABC\), \(b = 9\) cm and angle \(C = 30^\circ\). The area of the triangle is 18 cm\(^2\). Work out the length of \(a\). (3 marks)

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    Model answer

    \(18 = \dfrac{1}{2} \times a \times 9 \times \sin 30^\circ = \dfrac{9a}{4}\), so \(a = 18 \times \dfrac{4}{9} = 8\) cm.

    Mark scheme

    • \(18 = \dfrac{1}{2} \times a \times 9 \times \sin 30^\circ\) — M1
    • \(18 = \dfrac{9a}{4}\) or equivalent — M1
    • \(8\) — A1
  4. 4 Work out [3 marks]

    Diagram NOT accurately drawn. The diagram shows a triangular plot of land \(PQR\). Work out the exact area of the plot. (3 marks)

    A triangle diagram showing a triangular plot of land PQR.
    Show answerHide answer

    Model answer

    Area \(= \dfrac{1}{2} \times 10 \times 12 \times \sin 120^\circ = 60 \times \dfrac{\sqrt{3}}{2} = 30\sqrt{3}\) m\(^2\).

    Mark scheme

    • \(\dfrac{1}{2} \times 10 \times 12 \times \sin 120^\circ\) — M1
    • \(\sin 120^\circ = \dfrac{\sqrt{3}}{2}\) — M1
    • \(30\sqrt{3}\) — A1
  5. 5 Work out [5 marks]

    Diagram NOT accurately drawn. The diagram shows a circle, centre \(O\), with radius 4 cm. \(A\) and \(B\) are points on the circle and angle \(AOB = 120^\circ\). Work out the exact area of the segment bounded by the chord \(AB\) and the minor arc \(AB\). (5 marks)

    A circle with a chord AB and two radii enclosing an angle at the centre O, with the radius given.
    Show answerHide answer

    Model answer

    Sector \(= \dfrac{120}{360} \times \pi \times 4^2 = \dfrac{16\pi}{3}\). Triangle \(= \dfrac{1}{2} \times 4 \times 4 \times \sin 120^\circ = 8 \times \dfrac{\sqrt{3}}{2} = 4\sqrt{3}\). Segment \(= \dfrac{16\pi}{3} - 4\sqrt{3}\) cm\(^2\).

    Mark scheme

    • \(\dfrac{120}{360} \times \pi \times 4^2\) — M1
    • \(\dfrac{16\pi}{3}\) — A1
    • \(\dfrac{1}{2} \times 4 \times 4 \times \sin 120^\circ\) — M1
    • \(4\sqrt{3}\) — A1
    • \(\dfrac{16\pi}{3} - 4\sqrt{3}\) — A1
  6. 6 Work out [4 marks]

    A regular hexagon has sides of length 4 cm. Work out the exact area of the hexagon. (4 marks)

    Show answerHide answer

    Model answer

    The hexagon is made of 6 triangles with two sides of 4 cm and an angle of \(60^\circ\) between them. Each has area \(\dfrac{1}{2} \times 4 \times 4 \times \sin 60^\circ = 4\sqrt{3}\). So the total is \(6 \times 4\sqrt{3} = 24\sqrt{3}\) cm\(^2\).

    Mark scheme

    • Splits the hexagon into 6 triangles, each with an angle of \(60^\circ\) at the centre — M1
    • \(\dfrac{1}{2} \times 4 \times 4 \times \sin 60^\circ\) — M1
    • \(4\sqrt{3}\) for each triangle — A1
    • \(24\sqrt{3}\) — A1

Quick check

  1. 1

    What is the formula for the area of a triangle using a sine?

    1. A\(\dfrac{1}{2}ab\sin C\)
    2. B\(ab\sin C\)
    3. C\(\dfrac{1}{2}ab\cos C\)
    4. D\(\dfrac{1}{2}a\sin C\)
    Show answerHide answer

    A: \(\dfrac{1}{2}ab\sin C\)

    The area is half the product of two sides and the sine of the angle between them.

  2. 2

    Which angle is used in \(\dfrac{1}{2}ab\sin C\)?

    1. AThe angle opposite \(a\)
    2. BThe largest angle
    3. CThe right angle
    4. DThe angle between sides \(a\) and \(b\)
    Show answerHide answer

    D: The angle between sides \(a\) and \(b\)

    \(C\) is the included angle.

  3. 3

    What is \(\sin 90^\circ\)?

    1. A0
    2. B\(\dfrac{1}{2}\)
    3. C1
    4. D\(-1\)
    Show answerHide answer

    C: 1

    This is on the sine graph at its maximum.

  4. 4

    A triangle has sides 8 cm and 5 cm with an angle of \(30^\circ\) between them. What is the area?

    1. A20 cm\(^2\)
    2. B10 cm\(^2\)
    3. C\(20\sqrt{3}\) cm\(^2\)
    4. D40 cm\(^2\)
    Show answerHide answer

    B: 10 cm\(^2\)

    \(\dfrac{1}{2} \times 8 \times 5 \times \dfrac{1}{2} = 10\).

  5. 5

    A triangle has sides 6 cm and 8 cm with an angle of \(60^\circ\) between them. What is the exact area?

    1. A\(12\sqrt{3}\) cm\(^2\)
    2. B\(24\) cm\(^2\)
    3. C\(24\sqrt{3}\) cm\(^2\)
    4. D\(12\) cm\(^2\)
    Show answerHide answer

    A: \(12\sqrt{3}\) cm\(^2\)

    \(\dfrac{1}{2} \times 6 \times 8 \times \dfrac{\sqrt{3}}{2} = 12\sqrt{3}\).

  6. 6

    A triangle has sides of 10 cm and 8 cm and an area of 20 cm\(^2\). The included angle is acute. What is it?

    1. A\(60^\circ\)
    2. B\(45^\circ\)
    3. C\(150^\circ\)
    4. D\(30^\circ\)
    Show answerHide answer

    D: \(30^\circ\)

    \(20 = 40\sin C\), so \(\sin C = \dfrac{1}{2}\) and \(C = 30^\circ\).

  7. 7

    What is the area of a segment of a circle?

    1. AArea of the sector plus area of the triangle
    2. BArea of the circle minus the sector
    3. CArea of the sector minus area of the triangle
    4. DHalf the area of the sector
    Show answerHide answer

    C: Area of the sector minus area of the triangle

    The segment is the sector with the triangle removed.

  8. 8

    A sector of a circle with radius 6 cm has an angle of \(60^\circ\). What is the area of the sector?

    1. A\(36\pi\) cm\(^2\)
    2. B\(6\pi\) cm\(^2\)
    3. C\(12\pi\) cm\(^2\)
    4. D\(\pi\) cm\(^2\)
    Show answerHide answer

    B: \(6\pi\) cm\(^2\)

    \(\dfrac{60}{360} \times \pi \times 36 = 6\pi\).

  9. 9

    A triangle has two sides of 4 cm and 6 cm and an area of 6 cm\(^2\). What could the included angle be?

    1. A\(30^\circ\) or \(150^\circ\)
    2. B\(30^\circ\) only
    3. C\(60^\circ\) or \(120^\circ\)
    4. D\(45^\circ\) or \(135^\circ\)
    Show answerHide answer

    A: \(30^\circ\) or \(150^\circ\)

    \(6 = 12\sin C\), so \(\sin C = \dfrac{1}{2}\), which gives \(30^\circ\) or \(150^\circ\).

Trigonometry in 3D and Mixed Problems

Just this lesson
  1. 1 Work out [3 marks]

    Diagram NOT accurately drawn. The diagram shows a cuboid. Work out the length of the diagonal \(AG\). (3 marks)

    A cuboid with its three edge lengths labelled and a diagonal AG from one corner to the opposite corner.
    Show answerHide answer

    Model answer

    \(AG^2 = 4^2 + 4^2 + 7^2 = 16 + 16 + 49 = 81\), so \(AG = 9\) cm.

    Mark scheme

    • \(4^2 + 4^2 + 7^2\), or a base diagonal \(AC^2 = 4^2 + 4^2\) then \(AG^2 = AC^2 + 7^2\) — M1
    • \(81\) — M1
    • \(9\) — A1
  2. 2 Work out [4 marks]

    Diagram NOT accurately drawn. The diagram shows a cuboid. Work out the size of angle \(\theta\), the angle between \(AG\) and the base \(ABCD\). (4 marks)

    A cuboid with the base diagonal AC, the space diagonal AG and the angle theta between them.
    Show answerHide answer

    Model answer

    \(AC^2 = 6^2 + 8^2 = 100\), so \(AC = 10\) cm. In the right-angled triangle \(ACG\), \(\tan\theta = \dfrac{CG}{AC} = \dfrac{10}{10} = 1\), so \(\theta = 45^\circ\).

    Mark scheme

    • \(AC = 10\) — B1
    • \(\tan\theta = \dfrac{10}{10}\) — M1
    • \(\tan\theta = 1\) — A1
    • \(45\) — A1
  3. 3 Work out [5 marks]

    Diagram NOT accurately drawn. The diagram shows a pyramid with a square base \(ABCD\). The apex \(V\) is directly above the centre \(M\) of the base. All the edges are 6 cm long. (a) Work out the exact height \(VM\) of the pyramid. (3 marks) (b) Work out the size of angle \(\theta\), the angle between \(VA\) and the base. (2 marks)

    A pyramid with a square base and equal edges, with the vertical height VM and the angle theta between the edge VA and the base.
    Show answerHide answer

    Model answer

    (a) \(AC^2 = 6^2 + 6^2 = 72\), so \(AM = \dfrac{1}{2}AC = 3\sqrt{2}\). \(VM^2 = 6^2 - (3\sqrt{2})^2 = 36 - 18 = 18\), so \(VM = 3\sqrt{2}\) cm. (b) \(\cos\theta = \dfrac{AM}{VA} = \dfrac{3\sqrt{2}}{6} = \dfrac{\sqrt{2}}{2}\), so \(\theta = 45^\circ\).

    Mark scheme

    • (a) \(AM = 3\sqrt{2}\) — M1
    • (a) \(VM^2 = 6^2 - 18\) — M1
    • (a) \(3\sqrt{2}\) — A1
    • (b) \(\cos\theta = \dfrac{3\sqrt{2}}{6}\) or \(\tan\theta = 1\) — M1
    • (b) \(45\) — A1
  4. 4 Work out [4 marks]

    A ship sails 50 km from \(P\) to \(Q\) on a bearing of \(060^\circ\). It then sails 80 km from \(Q\) to \(R\) on a bearing of \(180^\circ\). Work out the distance \(PR\). (4 marks)

    Show answerHide answer

    Model answer

    The bearing of \(P\) from \(Q\) is \(240^\circ\), so angle \(PQR = 240 - 180 = 60^\circ\). \(PR^2 = 50^2 + 80^2 - 2 \times 50 \times 80 \times \cos 60^\circ = 2500 + 6400 - 4000 = 4900\), so \(PR = 70\) km.

    Mark scheme

    • Angle \(PQR = 60^\circ\) — M1
    • \(PR^2 = 50^2 + 80^2 - 2 \times 50 \times 80 \times \cos 60^\circ\) — M1
    • \(4900\) — A1
    • \(70\) — A1
  5. 5 Work out [5 marks]

    \(BT\) is a vertical tower on horizontal ground. The angle of elevation of \(T\) from a point \(A\) is \(30^\circ\), and from a point \(C\) is \(45^\circ\). Angle \(ABC = 90^\circ\) and \(AC = 80\) m. Work out the height of the tower. (5 marks)

    Show answerHide answer

    Model answer

    Let \(BT = h\). \(BA = \dfrac{h}{\tan 30^\circ} = h\sqrt{3}\) and \(BC = \dfrac{h}{\tan 45^\circ} = h\). In the right-angled triangle \(ABC\), \(AC^2 = 3h^2 + h^2 = 4h^2 = 6400\), so \(h^2 = 1600\) and \(h = 40\) m.

    Mark scheme

    • \(BA = h\sqrt{3}\), or \(\tan 30^\circ = \dfrac{h}{BA}\) — B1
    • \(BC = h\) — B1
    • \(AC^2 = BA^2 + BC^2 = 4h^2\) — M1
    • \(4h^2 = 6400\) — M1
    • \(40\) — A1
  6. 6 Show that [3 marks]

    \(ABCDEFGH\) is a cube with edges of length 4 cm. Show that triangle \(ACF\) is equilateral, and write down the size of angle \(CAF\). (3 marks)

    Show answerHide answer

    Model answer

    \(AC\), \(AF\) and \(CF\) are diagonals of faces of the cube. \(AC^2 = 4^2 + 4^2 = 32\), so \(AC = 4\sqrt{2}\). The other two face diagonals are the same length, so the triangle is equilateral and angle \(CAF = 60^\circ\).

    Mark scheme

    • \(AC^2 = 4^2 + 4^2 = 32\), so \(AC = 4\sqrt{2}\) — M1
    • \(AF\) and \(CF\) are also face diagonals, so all three sides are \(4\sqrt{2}\) — B1
    • Equilateral, so angle \(CAF = 60^\circ\) — C1

Quick check

  1. 1

    What is the formula for the space diagonal of a cuboid?

    1. A\(l + w + h\)
    2. B\(\sqrt{l^2 + w^2 + h^2}\)
    3. C\(\sqrt{l^2 + w^2}\)
    4. D\(l^2 + w^2 + h^2\)
    Show answerHide answer

    B: \(\sqrt{l^2 + w^2 + h^2}\)

    Use Pythagoras' theorem twice.

  2. 2

    A rectangle is 4 cm by 3 cm. What is the length of its diagonal?

    1. A5 cm
    2. B7 cm
    3. C\(\sqrt{7}\) cm
    4. D12 cm
    Show answerHide answer

    A: 5 cm

    \(\sqrt{16 + 9} = 5\).

  3. 3

    Where is the apex of a square-based pyramid?

    1. AAbove one of the corners
    2. BAbove the midpoint of one edge
    3. CAnywhere above the base
    4. DDirectly above the centre of the base
    Show answerHide answer

    D: Directly above the centre of the base

    For a right pyramid the apex is above the centre.

  4. 4

    A cuboid has edges 2 cm, 3 cm and 6 cm. What is the length of the space diagonal?

    1. A11 cm
    2. B\(\sqrt{11}\) cm
    3. C7 cm
    4. D5 cm
    Show answerHide answer

    C: 7 cm

    \(\sqrt{4 + 9 + 36} = \sqrt{49} = 7\).

  5. 5

    What is the angle between a line and a plane?

    1. AThe angle between the line and the vertical
    2. BThe angle between the line and its shadow on the plane
    3. CThe angle between two edges
    4. DAlways \(90^\circ\)
    Show answerHide answer

    B: The angle between the line and its shadow on the plane

    The shadow of the line on the plane is used.

  6. 6

    A cuboid has a base 4 cm by 3 cm and a height of 5 cm. What is \(\tan\theta\), where \(\theta\) is the angle between the diagonal \(AG\) and the base?

    1. A1
    2. B\(\dfrac{5}{3}\)
    3. C\(\dfrac{4}{5}\)
    4. D\(\dfrac{3}{5}\)
    Show answerHide answer

    A: 1

    The base diagonal is 5 cm and the height is 5 cm, so \(\tan\theta = \dfrac{5}{5} = 1\).

  7. 7

    A cube has edges of length 2 cm. What is the length of its space diagonal?

    1. A\(2\sqrt{2}\) cm
    2. B6 cm
    3. C\(4\sqrt{3}\) cm
    4. D\(2\sqrt{3}\) cm
    Show answerHide answer

    D: \(2\sqrt{3}\) cm

    \(\sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3}\).

  8. 8

    A square-based pyramid has a base with side 6 cm and a height of \(3\sqrt{2}\) cm. What is the angle between a sloping edge and the base?

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

    C: \(45^\circ\)

    Half the base diagonal is \(3\sqrt{2}\), the same as the height, so \(\tan\theta = 1\).

  9. 9

    In a 3D problem, which triangle contains the angle between the space diagonal of a cuboid and the base?

    1. AA triangle with the three edges at one corner
    2. BA right-angled triangle with the base diagonal, the vertical edge and the space diagonal
    3. CA triangle on the base only
    4. DA triangle on one face
    Show answerHide answer

    B: A right-angled triangle with the base diagonal, the vertical edge and the space diagonal

    The vertical edge is perpendicular to the base, so the triangle is right-angled.