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Date	May 2021	Marks available	2	Reference code	21M.1.SL.TZ2.4
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 2
Command term	Find	Question number	4	Adapted from	N/A

Question

The diagram below shows a helicopter hovering at point $H$ , $380 m$ vertically above a lake. Point $A$ is the point on the surface of the lake, directly below the helicopter.

Minta is swimming at a constant speed in the direction of point $A$ . Minta observes the helicopter from point $C$ as she looks upward at an angle of $25 °$ . After $15$ minutes, Minta is at point $B$ and she observes the same helicopter at an angle of $40 °$ .

Write down the size of the angle of depression from $H$ to $C$ .

[1]

Find the distance from $A$ to $C$ .

[2]

Find the distance from $B$ to $C$ .

[3]

Find Minta’s speed, in metres per hour.

[1]

Markscheme

$25 °$ A1

[1 mark]

$AC = \frac{380}{\tan 25 °}$ OR $AC = \sqrt{{(\frac{380}{\sin 25 °})}^{2} - 380^{2}}$ OR $\frac{380}{\sin 25 °} = \frac{AC}{\sin 65 °}$ (M1)

$AC = 815 m (814.912 \dots)$ A1

[2 marks]

METHOD 1

attempt to find $AB$ (M1)

$AB = \frac{380}{\tan 40 °}$

$= 453 m (452.866 \dots)$ (A1)

$BC = 814.912 \dots - 452.866 \dots$

$= 362 m (362.046 \dots)$ A1

METHOD 2

attempt to find $HB$ (M1)

$HB = \frac{380}{\sin 40 °}$

$591 m (= 591.175 \dots)$ (A1)

$BC = \frac{591.175 \dots \times \sin 15 °}{\sin 25 °}$

$= 362 m (362.046 \dots)$ A1

[3 marks]

$362.046 \dots \times 4$

$= 1450 {m h}^{- 1} (1448.18 \dots)$ A1

[1 mark]

Examiners report

[N/A]

Syllabus sections

Topic 3—Geometry and trigonometry » SL 3.3—Angles of elevation and depression

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Topic 3—Geometry and trigonometry

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