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Date	November Example question	Marks available	6	Reference code	EXN.1.SL.TZ0.11
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 0
Command term	Justify and Find	Question number	11	Adapted from	N/A

Question

A farmer owns a triangular field $ABC$ . The length of side $[AB]$ is $85 m$ and side $[AC]$ is $110 m$ . The angle between these two sides is $55 °$ .

Find the area of the field.

[3]

a.

The farmer would like to divide the field into two equal parts by constructing a straight fence from $A$ to a point $D$ on $[BC]$ .

Find $BD$ . Fully justify any assumptions you make.

[6]

b.

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

Area $= \frac{1}{2} \times 110 \times 85 \times \sin 55 °$ (M1)(A1)

$= 3830 (3829.53 \dots) m^{2}$ A1

Note: units must be given for the final A1 to be awarded.

[3 marks]

a.

${BC}^{2} = 110^{2} + 85^{2} - 2 \times 110 \times 85 \times \cos 55 °$ (M1)A1

$BC = 92.7 (92.7314 \dots) (m)$ A1

METHOD 1

Because the height and area of each triangle are equal they must have the same length base R1

$D$ must be placed half-way along $BC$ A1

$BD = \frac{92.731 \dots}{2} \approx 46.4 (m)$ A1

Note: the final two marks are dependent on the R1 being awarded.

METHOD 2

Let $C \hat{B} A = θ °$

$\frac{\sin θ}{110} = \frac{\sin 55 °}{92.731 \dots}$ M1

$\Rightarrow θ = 76.3 ° (76.3354 \dots)$

Use of area formula

$\frac{1}{2} \times 85 \times BD \times \sin (76.33 \dots °) = \frac{3829.53 \dots}{2}$ A1

$BD = 46.4 (46.365 \dots) (m)$ A1

[6 marks]

b.

Examiners report

[N/A]

a.

[N/A]

b.

Syllabus sections

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

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Topic 3—Geometry and trigonometry