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Date	May 2022	Marks available	3	Reference code	22M.2.SL.TZ2.1
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 2
Command term	Find	Question number	1	Adapted from	N/A

Question

The following diagram shows a circle with centre $O$ and radius $5$ metres.

Points $A$ and $B$ lie on the circle and $A \hat{O} B = 1.9$ radians.

Find the length of the chord $[AB]$ .

[3]

Find the area of the shaded sector.

[3]

Markscheme

EITHER

uses the cosine rule (M1)

${AB}^{2} = 5^{2} + 5^{2} - 2 \times 5 \times 5 \times \cos 1.9$ (A1)

uses right-angled trigonometry (M1)

$\frac{\frac{AB}{2}}{5} \sin 0.95$ (A1)

uses the sine rule (M1)

$α = \frac{1}{2} (π - 1.9) (= 0.6207 \dots)$

$\frac{AB}{\sin 1.9} = \frac{5}{\sin 0.6207 \dots}$ (A1)

THEN

$AB = 8.13415 \dots$

$AB = 8.13 (m)$ A1

[3 marks]

let the shaded area be $A$

METHOD 1

attempt at finding reflex angle (M1)

$A \hat{O} B = 2 π - 1.9 (= 4.3831 \dots)$

substitution into area formula (A1)

$A = \frac{1}{2} \times 5^{2} \times 4.3831 \dots$ OR $(\frac{2 π - 1.9}{2 π}) \times π (5^{2})$

$= 54.7898 \dots$

$= 54.8 (m^{2})$ A1

METHOD 2

let the area of the circle be $A_{C}$ and the area of the unshaded sector be $A_{U}$

$A = A_{C} - A_{U}$ (M1)

$A = π \times 5^{2} - \frac{1}{2} \times 5^{2} \times 1.9 (= 78.5398 \dots - 23.75)$ (A1)

$= 54.7898 \dots$

$= 54.8 (m^{2})$ A1

[3 marks]

Examiners report

Most students used the cosine rule to correctly find AB in part (a), although many found the arc length instead of the chord.

Part (b) was generally correctly solved. Some candidates found the area of the unshaded region rather than the shaded one.

[N/A]

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.2—2d and 3d trig, sine rule, cosine rule, area

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