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Date	November 2020	Marks available	6	Reference code	20N.2.SL.TZ0.S_7
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Find	Question number	S_7	Adapted from	N/A

Question

The following diagram shows a circle with centre $O$ and radius $1 cm$ . Points $A$ and $B$ lie on the circumference of the circle and $A \hat{O} B = 2 θ$ , where $0 < θ < \frac{π}{2}$ .

The tangents to the circle at $A$ and $B$ intersect at point $C$ .

Show that $AC = \tan θ$ .

[1]

a.

Find the value of $θ$ when the area of the shaded region is equal to the area of sector $OADB$ .

[6]

b.

Markscheme

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

correct working for $AC$ (seen anywhere) A1

eg $\tan θ = \frac{AC}{OA}, \tan θ = \frac{AC}{1}$

$AC = \tan θ$ AG N0

[1 mark]

a.

METHOD 1 (working with half the areas)

area of triangle $OAC$ or triangle $OBC$ (A1)

eg $\frac{1}{2} \times 1 \times \tan θ$

correct sector area (A1)

eg $\frac{1}{2} \times θ \times (1^{2}), \frac{1}{2} θ$

correct approach using their areas to find the shaded area (seen anywhere) (A1)

eg $A_{their triangle} - A_{their sector}, \frac{1}{2} θ - \frac{1}{2} \tan θ$

correct equation A1

eg $\frac{1}{2} \tan θ - \frac{1}{2} θ = \frac{1}{2} θ, \tan θ = 2 θ$

$1.16556$

$1.17$ A2 N4

METHOD 2 (working with entire kite and entire sector)

area of kite $OACB$ (A1)

eg $2 \times \frac{1}{2} \times 1 \times \tan θ, \frac{1}{2} \times \frac{1}{\cos θ} \times 2 \sin θ$

correct sector area (A1)

eg $\frac{1}{2} \times 2 θ \times (1^{2}), θ$

correct approach using their areas to find the shaded area (seen anywhere) (A1)

eg $A_{kite OACB} - A_{sector OADB}, θ - \tan θ$

correct equation A1

eg $\tan θ - θ = θ, \tan θ = 2 θ$

$1.16556$

$1.17$ A2 N4

[6 marks]

b.

Examiners report

[N/A]

a.

[N/A]

b.

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.4—Circle: radians, arcs, sectors

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Topic 3— Geometry and trigonometry