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Date	November 2020	Marks available	3	Reference code	20N.1.SL.TZ0.S_2
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 0
Command term	Find	Question number	S_2	Adapted from	N/A

Question

The following diagram shows a triangle $ABC$ .

$AC = 15 cm, BC = 10 cm$ , and $A \hat{B} C = θ$ .

Let $\sin C \hat{A} B = \frac{\sqrt{3}}{3}$ .

Given that $A \hat{B} C$ is acute, find $\sin θ$ .

[3]

Find $\cos (2 \times C \hat{A} B)$ .

[3]

Markscheme

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

METHOD 1 – (sine rule)

evidence of choosing sine rule (M1)

eg $\frac{\sin \hat{A}}{a} = \frac{\sin \hat{B}}{b}$

correct substitution (A1)

eg $\frac{\frac{\sqrt{3}}{3}}{10} = \frac{\sin θ}{15}, \frac{\sqrt{3}}{30} = \frac{\sin θ}{15}, \frac{\sqrt{3}}{30} = \frac{\sin B}{15}$

$\sin θ = \frac{\sqrt{3}}{2}$ A1 N2

METHOD 2 – (perpendicular from vertex $C$ )

valid approach to find perpendicular length (may be seen on diagram) (M1)

eg , $\frac{h}{15} = \frac{\sqrt{3}}{3}$

correct perpendicular length (A1)

eg $\frac{15 \sqrt{3}}{3}, 5 \sqrt{3}$

$\sin θ = \frac{\sqrt{3}}{2}$ A1 N2

Note: Do not award the final A mark if candidate goes on to state $\sin θ = \frac{π}{3}$ , as this demonstrates a lack of understanding.

[3 marks]

attempt to substitute into double-angle formula for cosine (M1)

$1 - 2 {(\frac{\sqrt{3}}{3})}^{2}, 2 {(\frac{\sqrt{6}}{3})}^{2} - 1, {(\frac{\sqrt{6}}{3})}^{2} - {(\frac{\sqrt{3}}{3})}^{2}, \cos (2 θ) = 1 - 2 {(\frac{\sqrt{3}}{2})}^{2}, 1 - 2 \sin^{2} (\frac{\sqrt{3}}{3})$

correct working (A1)

eg $1 - 2 \times \frac{3}{9}, 2 \times \frac{6}{9} - 1, \frac{6}{9} - \frac{3}{9}$

$\cos (2 \times C \hat{A} B) = \frac{3}{9} (= \frac{1}{3})$ A1 N2

[3 marks]

Examiners report

[N/A]

Syllabus sections

Topic 3—Geometry and trigonometry » SL 3.2—2d and 3d trig

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Topic 3—Geometry and trigonometry

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