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Date	November 2016	Marks available	3	Reference code	16N.2.AHL.TZ0.H_9
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Hence	Question number	H_9	Adapted from	N/A

Question

The diagram shows two circles with centres at the points A and B and radii $2r$ and $r$ , respectively. The point B lies on the circle with centre A. The circles intersect at the points C and D.

N16/5/MATHL/HP2/ENG/TZ0/09

Let $\alpha$ be the measure of the angle CAD and $\theta$ be the measure of the angle CBD in radians.

Find an expression for the shaded area in terms of $\alpha$ , $\theta$ and $r$ .

[3]

a.

Show that $\alpha = 4\arcsin \frac{1}{4}$ .

[2]

b.

Hence find the value of $r$ given that the shaded area is equal to 4.

[3]

c.

Markscheme

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

$A = 2(\alpha - \sin \alpha ){r^2} + \frac{1}{2}(\theta - \sin \theta ){r^2}$ M1A1A1

Note: Award M1A1A1 for alternative correct expressions eg. $A = 4\left( {\frac{\alpha }{2} - \sin \frac{\alpha }{2}} \right){r^2} + \frac{1}{2}\theta {r^2}$ .

[3 marks]

a.

METHOD 1

consider for example triangle ADM where M is the midpoint of BD M1

$\sin \frac{\alpha }{4} = \frac{1}{4}$ A1

$\frac{\alpha }{4} = \arcsin \frac{1}{4}$

$\alpha = 4\arcsin \frac{1}{4}$ AG

METHOD 2

attempting to use the cosine rule (to obtain $1 - \cos \frac{\alpha }{2} = \frac{1}{8}$ ) M1

$\sin \frac{\alpha }{4} = \frac{1}{4}$ (obtained from $\sin \frac{\alpha }{4} = \sqrt {\frac{{1 - \cos \frac{\alpha }{2}}}{2}}$ ) A1

$\frac{\alpha }{4} = \arcsin \frac{1}{4}$

$\alpha = 4\arcsin \frac{1}{4}$ AG

METHOD 3

$\sin \left( {\frac{\pi }{2} - \frac{\alpha }{4}} \right) = 2\sin \frac{\alpha }{2}$ where $\frac{\theta }{2} = \frac{\pi }{2} - \frac{\alpha }{4}$

$\cos \frac{\alpha }{4} = 4\sin \frac{\alpha }{4}\cos \frac{\alpha }{4}$ M1

Note: Award M1 either for use of the double angle formula or the conversion from sine to cosine.

$\frac{1}{4} = \sin \frac{\alpha }{4}$ A1

$\frac{\alpha }{4} = \arcsin \frac{1}{4}$

$\alpha = 4\arcsin \frac{1}{4}$ AG

[2 marks]

b.

(from triangle ADM), $\theta = \pi - \frac{\alpha }{2}{\text{ }}\left( { = \pi - 2\arcsin \frac{1}{4} = 2\arcsin \frac{1}{4} = 2.6362 \ldots } \right)$ A1

attempting to solve $2(\alpha - \sin \alpha ){r^2} + \frac{1}{2}(\theta - \sin \theta ){r^2} = 4$

with $\alpha = 4\arcsin \frac{1}{4}$ and $\theta = \pi - \frac{\alpha }{2}{\text{ }}\left( { = 2\arccos \frac{1}{4}} \right)$ for $r$ (M1)

$r = 1.69$ A1

[3 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 3— Geometry and trigonometry