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Date	November 2016	Marks available	5	Reference code	16N.2.AHL.TZ0.H_7
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Find	Question number	H_7	Adapted from	N/A

Question

In a triangle ${\text{ABC, AB}} = 4{\text{ cm, BC}} = 3{\text{ cm}}$ and ${\rm{B\hat AC}} = \frac{\pi }{9}$ .

Use the cosine rule to find the two possible values for AC.

[5]

Find the difference between the areas of the two possible triangles ABC.

[3]

Markscheme

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

METHOD 1

let ${\text{AC}} = x$

${3^2} = {x^2} + {4^2} - 8x\cos \frac{\pi }{9}$ M1A1

attempting to solve for $x$ (M1)

$x = 1.09,{\text{ }}6.43$ A1A1

METHOD 2

let ${\text{AC}} = x$

using the sine rule to find a value of $C$ M1

${4^2} = {x^2} + {3^2} - 6x\cos (152.869 \ldots ^\circ ) \Rightarrow x = 1.09$ (M1)A1

${4^2} = {x^2} + {3^2} - 6x\cos (27.131 \ldots ^\circ ) \Rightarrow x = 6.43$ (M1)A1

METHOD 3

let ${\text{AC}} = x$

using the sine rule to find a value of $B$ and a value of $C$ M1

obtaining $B = 132.869 \ldots ^\circ ,{\text{ }}7.131 \ldots ^\circ$ and $C = 27.131 \ldots ^\circ ,{\text{ }}152.869 \ldots ^\circ$ A1

$(B = 2.319 \ldots ,{\text{ }}0.124 \ldots$ and $C = 0.473 \ldots ,{\text{ }}2.668 \ldots )$

attempting to find a value of $x$ using the cosine rule (M1)

$x = 1.09,{\text{ }}6.43$ A1A1

Note: Award M1A0(M1)A1A0 for one correct value of $x$

[5 marks]

$\frac{1}{2} \times 4 \times 6.428 \ldots \times \sin \frac{\pi }{9}$ and $\frac{1}{2} \times 4 \times 1.088 \ldots \times \sin \frac{\pi }{9}$ (A1)

( $4.39747 \ldots$ and $0.744833 \ldots$ )

let $D$ be the difference between the two areas

$D = \frac{1}{2} \times 4 \times 6.428 \ldots \times \sin \frac{\pi }{9} - \frac{1}{2} \times 4 \times 1.088 \ldots \times \sin \frac{\pi }{9}$ (M1)

$(D = 4.39747 \ldots - 0.744833 \ldots )$

$= 3.65{\text{ (c}}{{\text{m}}^2})$ A1

[3 marks]

Examiners report

[N/A]

Syllabus sections

Topic 3—Geometry and trigonometry » AHL 3.7—Radians

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