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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 10cm. The clock’s hour hand has a length of 6cm.

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 10cm with minute and hour hands both of length 10cm. The time shown is 6:00 am. The bottom of the clock face is located 3cm above a horizontal bookshelf.

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

hθ=10cosθ+13, θ0,

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

gθ=-10cosθ12+13, θ0,

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

Find the size of angle AO^B in degrees.

[2]
a.

Find the distance between points A and B.

[3]
b.

Find the size of angle θ in degrees.

[2]
c.

Calculate the length of arc AC.

[2]
d.

Calculate the area of the shaded sector, AOC.

[2]
e.

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

[1]
f.

Find the value of h when θ=160°.

[2]
g.

Write down the amplitude of g(θ).

[1]
h.

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

Find the smallest possible value of k.

[2]
i.

Markscheme

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

120°                  A1


[2 marks]

a.

substitution in cosine rule                   (M1)

AB2=102+62-2×10×6×cos120°                  (A1)

AB=14 cm                  A1


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


[3 marks]

b.

θ=13×6                   (M1)

=78°                A1


[2 marks]

c.

substitution into the formula for arc length                 (M1)

l=78360×2×π×10  OR  l=13π30×10

=13.6 cm  13.6135, 4.33π, 13π3                A1


[2 marks]

d.

substitution into the area of a sector                (M1)

A=78360×π×102  OR  l=12×13π30×102

=68.1 cm2  68.0678, 21.7π, 65π3                A1


[2 marks]

e.

23              A1


[1 mark]

f.

correct substitution              (M1)

h=10cos160°+13

=3.60 cm  3.60307              A1


[2 marks]

g.

10           A1


[1 mark]

h.

EITHER

10×cosθ+13=-10×cosθ12+13              (M1)


OR

                             (M1)


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

 

THEN

k=196°  196.363          A1


Note: The answer 166.153 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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