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GCSE / IGCSE Maths – Trigonometry

SOHCAHTOA, sine and cosine rules, 3D trigonometry, trigonometric graphs.

12 questions  ·  56 marks  ·  answers included  ·  GCSE/IGCSE (Higher)
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This worksheet covers:

  • Right-angled trigonometry (SOHCAHTOA)
  • Exact values
  • The sine and cosine rules
  • Area of a triangle
  • 3D problems
  • Trigonometric graphs

Section A is non-calculator; a scientific calculator is needed for Sections B and C. Allow about 60 minutes.

Questions

Section A – Skills

Non-calculator. Give exact answers.
1

Write down the exact value of:(a) $\sin 30^\circ$(b) $\cos 45^\circ$(c) $\tan 60^\circ$(d) $\cos 90^\circ$

Show answer
(a) $\frac{1}{2}$(b) $\frac{\sqrt{2}}{2}$ or $\frac{1}{\sqrt{2}}$(c) $\sqrt{3}$(d) $0$
[4]
2

A right-angled triangle has a hypotenuse of $10$ cm and one angle of $30^\circ$. Find the exact lengths of the other two sides.

Show answer
Opposite $30^\circ$: $10 \sin 30^\circ = 5$ cm; adjacent: $10 \cos 30^\circ = 5\sqrt{3}$ cm
[3]
3

In triangle $PQR$, angle $Q = 90^\circ$, $PQ = 3$ cm and $QR = 4$ cm.(a) Find $PR$.(b) Write down the value of $\tan P$.(c) Write down the value of $\sin R$.

Show answer
(a) $PR = 5$ cm(b) $\tan P = \frac{4}{3}$(c) $\sin R = \frac{3}{5}$
[4]
4

The graph of $y = \sin x$ is drawn for $0^\circ \le x \le 360^\circ$.(a) Write down the maximum value of $y$ and the value of $x$ at which it occurs.(b) Write down the two values of $x$ in this range for which $\sin x = 0$, other than $x = 0^\circ$.(c) Given that $\sin 50^\circ \approx 0.766$, use the symmetry of the graph to write down another solution of $\sin x = 0.766$ in the range $0^\circ \le x \le 360^\circ$.

Show answer
(a) Maximum $y = 1$ at $x = 90^\circ$(b) $x = 180^\circ$ and $x = 360^\circ$(c) $x = 180^\circ – 50^\circ = 130^\circ$
[4]

Section B – Exam-style

Calculator allowed. Give answers to 3 significant figures unless stated.
5

(a) A right-angled triangle has hypotenuse $8$ cm and one angle of $35^\circ$. Find the length of the side opposite the $35^\circ$ angle.(b) In a different right-angled triangle, the side adjacent to an angle of $52^\circ$ is $5$ cm. Find the hypotenuse.(c) A ramp rises $7$ m over a horizontal distance of $12$ m. Find the angle the ramp makes with the horizontal.

Show answer
(a) $8 \sin 35^\circ = 4.59$ cm(b) $\frac{5}{\cos 52^\circ} = 8.12$ cm(c) $\tan^{-1}\left(\frac{7}{12}\right) = 30.3^\circ$
[6]
6

A ladder of length $6$ m leans against a vertical wall, making an angle of $70^\circ$ with the horizontal ground. Find how far the foot of the ladder is from the wall, giving your answer to 2 decimal places.

Show answer
$6 \cos 70^\circ = 2.05$ m
[2]
7

In triangle $ABC$, angle $A = 38^\circ$, angle $B = 71^\circ$ and $BC = 9$ cm.(a) Find the length of $AC$.(b) Find the size of angle $C$.

Show answer
(a) $\frac{AC}{\sin 71^\circ} = \frac{9}{\sin 38^\circ} \Rightarrow AC = 13.8$ cm(b) $C = 180^\circ – 38^\circ – 71^\circ = 71^\circ$
[4]
8

In triangle $XYZ$, $XY = 7$ cm, $XZ = 10$ cm and angle $X = 58^\circ$.(a) Find the length of $YZ$.(b) Find the area of triangle $XYZ$.

Show answer
(a) $YZ^2 = 7^2 + 10^2 – 2(7)(10)\cos 58^\circ \Rightarrow YZ = 8.65$ cm(b) $\frac{1}{2}(7)(10)\sin 58^\circ = 29.7$ cm$^2$
[5]
9

A triangle has sides of length $8$ cm, $11$ cm and $14$ cm. Find the size of the largest angle.

Show answer
$\cos\theta = \frac{8^2 + 11^2 – 14^2}{2(8)(11)} = -\frac{11}{176} \Rightarrow \theta = 93.6^\circ$
[3]
10

A cuboid has length $8$ cm, width $6$ cm and height $5$ cm. Vertex $A$ is at one bottom corner and vertex $G$ is the top corner diagonally opposite.(a) Find the length of the base diagonal $AC$.(b) Find the length of the space diagonal $AG$.(c) Find the angle that $AG$ makes with the base.

Show answer
(a) $AC = \sqrt{8^2 + 6^2} = 10$ cm(b) $AG = \sqrt{8^2 + 6^2 + 5^2} = \sqrt{125} = 11.2$ cm(c) $\tan^{-1}\left(\frac{5}{10}\right) = 26.6^\circ$
[6]

Section C – Challenge

Calculator allowed.
11

A ship sails $12$ km from port $P$ on a bearing of $040^\circ$ to point $Q$. It then sails $9$ km on a bearing of $150^\circ$ to point $R$.(a) Show that angle $PQR = 70^\circ$.(b) Find the distance $PR$.(c) Find the bearing of $R$ from $P$.

Show answer
(a) At $Q$ the reverse bearing of $P$ is $220^\circ$; $220^\circ – 150^\circ = 70^\circ$(b) $PR^2 = 12^2 + 9^2 – 2(12)(9)\cos 70^\circ \Rightarrow PR = 12.3$ km(c) $\sin(QPR) = \frac{9 \sin 70^\circ}{12.3} \Rightarrow QPR = 43.5^\circ$; bearing $= 040^\circ + 43.5^\circ = 083.5^\circ$ (accept $084^\circ$)
[7]
12

A square-based pyramid has base side $10$ cm and vertical height $9$ cm. The apex $V$ is directly above the centre $O$ of the base $ABCD$.(a) Find the length of $OA$.(b) Find the angle that edge $VA$ makes with the base.(c) $M$ is the midpoint of $AB$. Find the length $VM$ and hence the angle between the face $VAB$ and the base.

Show answer
(a) $OA = \frac{1}{2}\sqrt{10^2 + 10^2} = 5\sqrt{2} = 7.07$ cm(b) $\tan^{-1}\left(\frac{9}{5\sqrt{2}}\right) = 51.8^\circ$(c) $VM = \sqrt{9^2 + 5^2} = \sqrt{106} = 10.3$ cm; angle $= \tan^{-1}\left(\frac{9}{5}\right) = 60.9^\circ$
[8]

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