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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 OO. The clock’s minute hand has a length of 10cm10cm. The clock’s hour hand has a length of 6cm6cm.

At 4:004:00 pm the endpoint of the minute hand is at point AA and the endpoint of the hour hand is at point BB.

 

Between 4:004:00 pm and 4:134:13 pm, the endpoint of the minute hand rotates through an angle, θθ, from point AA to point CC. This is illustrated in the diagram.

A second clock is illustrated in the diagram below. The clock face has radius 10cm10cm with minute and hour hands both of length 10cm10cm. The time shown is 6:006:00 am. The bottom of the clock face is located 3cm3cm above a horizontal bookshelf.

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

h(θ)=10cosθ+13, θ0,h(θ)=10cosθ+13, θ0,

where θθ is the angle rotated by the minute hand from 6:006:00 am.

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

g(θ)=-10cos(θ12)+13, θ0,g(θ)=10cos(θ12)+13, θ0,

where θθ is the angle in degrees rotated by the minute hand from 6:006:00 am.

Find the size of angle AˆOBAˆOB in degrees.

[2]
a.

Find the distance between points AA and BB.

[3]
b.

Find the size of angle θθ in degrees.

[2]
c.

Calculate the length of arc ACAC.

[2]
d.

Calculate the area of the shaded sector, AOCAOC.

[2]
e.

Write down the height of the endpoint of the minute hand above the bookshelf at 6:006:00 am.

[1]
f.

Find the value of hh when θ=160°θ=160°.

[2]
g.

Write down the amplitude of g(θ)g(θ).

[1]
h.

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

Find the smallest possible value of kk.

[2]
i.

Markscheme

4×360°124×360°12  OR  4×30°4×30°                   (M1)

120°120°                  A1


[2 marks]

a.

substitution in cosine rule                   (M1)

AB2=102+62-2×10×6×cos(120°)AB2=102+622×10×6×cos(120°)                  (A1)

AB=14AB=14 cmcm                  A1


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


[3 marks]

b.

θ=13×6θ=13×6                   (M1)

=78°=78°                A1


[2 marks]

c.

substitution into the formula for arc length                 (M1)

l=78360×2×π×10l=78360×2×π×10  OR  l=13π30×10l=13π30×10

=13.6 cm  (13.6135, 4.33π, 13π3)=13.6 cm  (13.6135, 4.33π, 13π3)                A1


[2 marks]

d.

substitution into the area of a sector                (M1)

A=78360×π×102A=78360×π×102  OR  l=12×13π30×102l=12×13π30×102

=68.1 cm2  (68.0678, 21.7π, 65π3)=68.1 cm2  (68.0678, 21.7π, 65π3)                A1


[2 marks]

e.

2323              A1


[1 mark]

f.

correct substitution              (M1)

h=10cos(160°)+13h=10cos(160°)+13

=3.60 cm  (3.60307)=3.60 cm  (3.60307)              A1


[2 marks]

g.

1010           A1


[1 mark]

h.

EITHER

10×cos(θ)+13=-10×cos(θ12)+1310×cos(θ)+13=10×cos(θ12)+13              (M1)


OR

                             (M1)


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

 

THEN

k=196°  (196.363)k=196°  (196.363)          A1


Note: The answer 166.153166.153 is incorrect but the correct method is implicit. Award (M1)A0.


[2 marks]

i.

Examiners report

[N/A]
a.
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b.
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c.
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d.
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e.
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f.
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g.
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h.
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i.

Syllabus sections

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

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