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Date May 2019 Marks available 2 Reference code 19M.2.SL.TZ2.S_4
Level Standard Level Paper Paper 2 Time zone Time zone 2
Command term Find Question number S_4 Adapted from N/A

Question

OAB is a sector of the circle with centre O and radius r , as shown in the following diagram.

The angle AOB is θ radians, where  0 < θ < π 2 .

The point C lies on OA and OA is perpendicular to BC.

Show that OC = r cos θ .

[1]
a.

Find the area of triangle OBC in terms of r and θ.

[2]
b.

Given that the area of triangle OBC is  3 5  of the area of sector OAB, find θ.

[4]
c.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

cos θ = OC r      A1

OC = r cos θ   AG N0

[1 mark]

a.

valid approach    (M1)

eg    1 2 OC × OB sin θ ,   BC = r sin θ 1 2 r cos θ × BC ,   1 2 r sin θ × OC

area = 1 2 r 2 sin θ cos θ   ( = 1 4 r 2 sin ( 2 θ ) )   (must be in terms of r and θ)      A1 N2

[2 marks]

b.

valid attempt to express the relationship between the areas (seen anywhere)        (M1)

eg   OCB =  3 5 OBA ,   1 2 r 2 sin θ cos θ = 3 5 × 1 2 r 2 θ ,   1 4 r 2 sin 2 θ = 3 10 r 2 θ

correct equation in terms of θ only      A1

eg    sin θ cos θ = 3 5 θ ,   1 4 sin 2 θ = 3 10 θ

valid attempt to solve their equation        (M1)

eg    sketch,  −0.830017,  0

0.830017

θ = 0.830      A1 N2

Note: Do not award final A1 if additional answers given.

[4 marks]

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.4—Circle: radians, arcs, sectors
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Topic 2—Functions » SL 2.10—Solving equations graphically and analytically
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Topic 3— Geometry and trigonometry

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