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Date	May 2021	Marks available	2	Reference code	21M.2.SL.TZ1.2
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 1
Command term	Find	Question number	2	Adapted from	N/A

Question

The diagram below shows a circular clockface with centre $O$ . The clock’s minute hand has a length of $10 cm$ . The clock’s hour hand has a length of $6 cm$ .

At $4 : 00$ pm the endpoint of the minute hand is at point $A$ and the endpoint of the hour hand is at point $B$ .

Between $4 : 00$ pm and $4 : 13$ pm, the endpoint of the minute hand rotates through an angle, $θ$ , from point $A$ to point $C$ . This is illustrated in the diagram.

A second clock is illustrated in the diagram below. The clock face has radius $10 cm$ with minute and hour hands both of length $10 cm$ . The time shown is $6 : 00$ am. The bottom of the clock face is located $3 cm$ above a horizontal bookshelf.

The height, $h$ centimetres, of the endpoint of the minute hand above the bookshelf is modelled by the function

$h (θ) = 10 \cos θ + 13, θ \geq 0,$

where $θ$ is the angle rotated by the minute hand from $6 : 00$ am.

The height, $g$ centimetres, of the endpoint of the hour hand above the bookshelf is modelled by the function

$g (θ) = - 10 \cos (\frac{θ}{12}) + 13, θ \geq 0,$

where $θ$ is the angle in degrees rotated by the minute hand from $6 : 00$ am.

Find the size of angle $A \hat{O} B$ in degrees.

[2]

Find the distance between points $A$ and $B$ .

[3]

Find the size of angle $θ$ in degrees.

[2]

Calculate the length of arc $AC$ .

[2]

Calculate the area of the shaded sector, $AOC$ .

[2]

Write down the height of the endpoint of the minute hand above the bookshelf at $6 : 00$ am.

[1]

Find the value of $h$ when $θ = 160 °$ .

[2]

Write down the amplitude of $g (θ)$ .

[1]

The endpoints of the minute hand and hour hand meet when $θ = k$ .

Find the smallest possible value of $k$ .

[2]

Markscheme

$4 \times \frac{360 °}{12}$ OR $4 \times 30 °$ (M1)

$120 °$ A1

[2 marks]

substitution in cosine rule (M1)

${AB}^{2} = 10^{2} + 6^{2} - 2 \times 10 \times 6 \times \cos (120 °)$ (A1)

$AB = 14$ $cm$ A1

Note: Follow through marks in part (b) are contingent on working seen.

[3 marks]

$θ = 13 \times 6$ (M1)

$= 78 °$ A1

[2 marks]

substitution into the formula for arc length (M1)

$l = \frac{78}{360} \times 2 \times π \times 10$ OR $l = \frac{13 π}{30} \times 10$

$= 13.6 cm (13.6135 \dots, 4.33 π, \frac{13 π}{3})$ A1

[2 marks]

substitution into the area of a sector (M1)

$A = \frac{78}{360} \times π \times 10^{2}$ OR $l = \frac{1}{2} \times \frac{13 π}{30} \times 10^{2}$

$= 68.1 {cm}^{2} (68.0678 \dots, 21.7 π, \frac{65 π}{3})$ A1

[2 marks]

$23$ A1

[1 mark]

correct substitution (M1)

$h = 10 \cos (160 °) + 13$

$= 3.60 cm (3.60307 \dots)$ A1

[2 marks]

$10$ A1

[1 mark]

EITHER

$10 \times \cos (θ) + 13 = - 10 \times \cos (\frac{θ}{12}) + 13$ (M1)

(M1)

Note: Award M1 for equating the functions. Accept a sketch of $h (θ)$ and $g (θ)$ with point(s) of intersection marked.

THEN

$k = 196 ° (196.363 \dots)$ A1

Note: The answer $166.153 \dots$ is incorrect but the correct method is implicit. Award (M1)A0.

[2 marks]

Examiners report

[N/A]

Syllabus sections

Topic 3—Geometry and trigonometry » SL 3.2—2d and 3d trig

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21M.1.SL.TZ1.9:
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21M.2.SL.TZ2.5a:
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21N.1.AHL.TZ0.11:
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21N.2.SL.TZ0.4d:
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Topic 3—Geometry and trigonometry

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