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Date	November 2021	Marks available	5	Reference code	21N.2.SL.TZ0.3
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Find	Question number	3	Adapted from	N/A

Question

Consider a triangle $ABC$ , where $AC = 12, CB = 7$ and $B \hat{A} C = 25 °$ .

Find the smallest possible perimeter of triangle $ABC$ .

Markscheme

EITHER

attempt to use cosine rule (M1)

$12^{2} + {AB}^{2} - 2 \times 12 \times \cos 25 ° \times AB = 7^{2}$ OR ${AB}^{2} - 21.7513 \dots AB + 95 = 0$ (A1)

at least one correct value for $AB$ (A1)

$AB = 6.05068 \dots$ OR $AB = 15.7007 \dots$

using their smaller value for $AB$ to find minimum perimeter (M1)

$12 + 7 + 6.05068 \dots$

attempt to use sine rule (M1)

$\frac{\sin B}{12} = \frac{\sin 25 °}{7}$ OR $\sin B = 0.724488 \dots$ OR $\hat{B} = 133.573 \dots °$ OR $\hat{B} = 46.4263 \dots °$ (A1)

at least one correct value for $C$ (A1)

$\hat{C} = 21.4263 \dots °$ OR $\hat{C} = 108.573 \dots °$

using their acute value for $\hat{C}$ to find minimum perimeter (M1)

$12 + 7 + \sqrt{12^{2} + 7^{2} - 2 \times 12 \times 7 \cos 21.4263 \dots °}$ OR $12 + 7 + \frac{7 \sin 21.4263 \dots °}{\sin 25 °}$

THEN

$25.0506 \dots$

minimum perimeter $= 25.1$ . A1

[5 marks]

Examiners report

[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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