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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 $r$ $r$ , as shown in the following diagram.

The angle AOB is $\theta$ $\theta$ radians, where $0 < \theta < \frac{\pi }{2}$ $0 < \theta < \frac{\pi }{2}$ .

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

Show that ${\text{OC}} = r\,{\text{cos}}\,\theta$ ${\text{OC}} = r\,{\text{cos}}\,\theta$ .

[1]

a.

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

[2]

b.

Given that the area of triangle OBC is $\frac{3}{5}$ $\frac{3}{5}$ 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.

${\text{cos}}\,\theta = \frac{{{\text{OC}}}}{r}$ ${\text{cos}}\,\theta = \frac{{{\text{OC}}}}{r}$ A1

${\text{OC}} = r\,{\text{cos}}\,\theta$ ${\text{OC}} = r\,{\text{cos}}\,\theta$ AG N0

[1 mark]

a.

valid approach (M1)

eg $\frac{1}{2}{\text{OC}} \times {\text{OB}}\,{\text{sin}}\,\theta$ $\frac{1}{2}{\text{OC}} \times {\text{OB}}\,{\text{sin}}\,\theta$ , ${\text{BC}} = r\,{\text{sin}}\,\theta$ ${\text{BC}} = r\,{\text{sin}}\,\theta$ , $\frac{1}{2}r\,{\text{cos}}\,\theta \times {\text{BC}}$ $\frac{1}{2}r\,{\text{cos}}\,\theta \times {\text{BC}}$ , $\frac{1}{2}r\,{\text{sin}}\,\theta \times {\text{OC}}$ $\frac{1}{2}r\,{\text{sin}}\,\theta \times {\text{OC}}$

area $= \frac{1}{2}{r^2}\,{\text{sin}}\,\theta \,{\text{cos}}\,\theta$ $= \frac{1}{2}{r^2}\,{\text{sin}}\,\theta \,{\text{cos}}\,\theta$ $\left( { = \frac{1}{4}{r^2}\,{\text{sin}}\left( {2\theta } \right)} \right)$ $\left( { = \frac{1}{4}{r^2}\,{\text{sin}}\left( {2\theta } \right)} \right)$ (must be in terms of $r$ $r$ and θ) A1 N2

[2 marks]

b.

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

eg OCB = $\frac{3}{5}$ $\frac{3}{5}$ OBA , $\frac{1}{2}{r^2}\,{\text{sin}}\,\theta \,{\text{cos}}\,\theta = \frac{3}{5} \times \frac{1}{2}{r^2}\theta$ $\frac{1}{2}{r^2}\,{\text{sin}}\,\theta \,{\text{cos}}\,\theta = \frac{3}{5} \times \frac{1}{2}{r^2}\theta$ , $\frac{1}{4}{r^2}\,{\text{sin}}\,2\theta = \frac{3}{{10}}{r^2}\theta$ $\frac{1}{4}{r^2}\,{\text{sin}}\,2\theta = \frac{3}{{10}}{r^2}\theta$

correct equation in terms of θ only A1

eg ${\text{sin}}\,\theta \,{\text{cos}}\,\theta = \frac{3}{5}\theta$ ${\text{sin}}\,\theta \,{\text{cos}}\,\theta = \frac{3}{5}\theta$ , $\frac{1}{4}{\text{sin}}\,2\theta = \frac{3}{{10}}\theta$ $\frac{1}{4}{\text{sin}}\,2\theta = \frac{3}{{10}}\theta$

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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