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Date	May 2021	Marks available	2	Reference code	21M.2.SL.TZ2.7
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 2
Command term	Find	Question number	7	Adapted from	N/A

Question

The six blades of a windmill rotate around a centre point $C$ . Points $A$ and $B$ and the base of the windmill are on level ground, as shown in the following diagram.

From point $A$ the angle of elevation of point $C$ is $0.6$ radians.

An observer walks $7$ metres from point $A$ to point $B$ .

The observer keeps walking until he is standing directly under point $C$ . The observer has a height of $1.8$ metres, and as the blades of the windmill rotate, the end of each blade passes $2.5$ metres over his head.

One of the blades is painted a different colour than the others. The end of this blade is labelled point $D$ . The height $h$ , in metres, of point $D$ above the ground can be modelled by the function $h (t) = p \cos (\frac{3 π}{10} t) + q$ , where $t$ is in seconds and $p, q \in ℝ$ . When $t = 0$ , point $D$ is at its maximum height.

Given that point $A$ is $12$ metres from the base of the windmill, find the height of point $C$ above the ground.

[2]

Find the angle of elevation of point $C$ from point $B$ .

[2]

Find the length of each blade of the windmill.

[2]

Find the value of $p$ and the value of $q$ .

[4]

If the observer stands directly under point $C$ for one minute, point $D$ will pass over his head $n$ times.

Find the value of $n$ .

[3]

Markscheme

$\tan 0.6 = \frac{h}{12}$ (M1)

$8.20964 \dots$

$8.21 (m)$ A1

[2 marks]

$\tan B = \frac{8.2096 \dots}{5}$ OR $\tan^{- 1} 1.6419 \dots$ (A1)

$1.02375 \dots$

$1.02$ (radians) (accept $58.7 °$ ) A1

[2 marks]

$x + 1.8 + 2.5 = 8.20964 \dots$ (or equivalent) (A1)

$3.90964 \dots$

$3.91$ ( $m$ ) A1

[2 marks]

METHOD 1

recognition that blade length = amplitude, $p = \frac{max - min}{2}$ (M1)

$p = 3.91$ A1

centre of windmill = vertical shift, $q = \frac{max + min}{2}$ (M1)

$q = 8.21$ A1

METHOD 2

attempting to form two equations in terms of $p$ and $q$ (M1)(M1)

$12.1192 \dots = p \cos (\frac{3 π}{10} \cdot 0) + q, 4.3000 \dots = p \cos (\frac{3 π}{10} \cdot \frac{10}{3}) + q$

$p = 3.91$ A1

$q = 8.21$ A1

[4 marks]

appropriate working towards finding the period (M1)

$period = \frac{2 π}{\frac{3 π}{10}} (= 6.6666 \dots)$

rotations per minute $= \frac{60}{their period}$ (M1)

$n = 9$ (must be an integer) (accept $n = 10$ , $n = 18$ , $n = 19$ ) A1

[3 marks]

Examiners report

[N/A]

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.3—Applications: angles of elevation and depression, bearings

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Topic 3— Geometry and trigonometry

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