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Date	May 2021	Marks available	5	Reference code	21M.2.SL.TZ2.2
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 2
Command term	Find	Question number	2	Adapted from	N/A

Question

A farmer owns a field in the shape of a triangle $ABC$ $ABC$ such that $AB = 650 m, AC = 1005 m$ $AB = 650 m, AC = 1005 m$ and $BC = 1225 m$ $BC = 1225 m$ .

The local town is planning to build a highway that will intersect the borders of the field at points $D$ $D$ and $E$ $E$ , where $DC = 210 m$ $DC = 210 m$ and $CÊD = 100 °$ $CÊD = 100 °$ , as shown in the diagram below.

The town wishes to build a carpark here. They ask the farmer to exchange the part of the field represented by triangle $DCE$ $DCE$ . In return the farmer will get a triangle of equal area $ADF$ $ADF$ , where $F$ $F$ lies on the same line as $D$ $D$ and $E$ $E$ , as shown in the diagram above.

Find the size of $AĈB$ $AĈB$ .

[3]

Find $DE$ $DE$ .

[3]

Find the area of triangle $DCE$ $DCE$ .

[5]

Estimate $DF$ $DF$ . You may assume the highway has a width of zero.

[4]

Markscheme

use of cosine rule (M1)

$AĈB = {cos}^{- 1} (\frac{1005^{2} + 1225^{2} - 650^{2}}{2 \times 1005 \times 1225})$ $AĈB = \cos^{- 1} (\frac{1005^{2} + 1225^{2} - 650^{2}}{2 \times 1005 \times 1225})$ (A1)

$= 32 ° (31.9980 \dots)$ $= 32 ° (31.9980 \dots)$ A1

[3 marks]

use of sine rule (M1)

$\frac{DE}{sin 31.9980 \dots °} = \frac{210}{sin 100 °}$ $\frac{DE}{\sin 31.9980 \dots °} = \frac{210}{\sin 100 °}$ (A1)

$(DE =) 113 m (112.9937 \dots)$ $(DE =) 113 m (112.9937 \dots)$ A1

[3 marks]

METHOD 1

$180 ° - (100 ° + their part (a))$ $180 ° - (100 ° + their part (a))$ (M1)

$= 48.0019 \dots °$ $= 48.0019 \dots °$ OR $0.837791 \dots$ $0.837791 \dots$ (A1)

substituted area of triangle formula (M1)

$\frac{1}{2} \times 112.9937 \dots \times 210 \times sin 48.002 °$ $\frac{1}{2} \times 112.9937 \dots \times 210 \times \sin 48.002 °$ (A1)

$8820 m^{2} (8817.18 \dots)$ $8820 m^{2} (8817.18 \dots)$ A1

METHOD 2

$\frac{CE}{sin (180 - 100 - their part (a))} = \frac{210}{sin 100}$ $\frac{CE}{\sin (180 - 100 - their part (a))} = \frac{210}{\sin 100}$ (M1)

$(CE =) 158.472 \dots$ $(CE =) 158.472 \dots$ (A1)

substituted area of triangle formula (M1)

EITHER

$\frac{1}{2} \times 112.993 \dots \times 158.472 \dots \times sin 100$ $\frac{1}{2} \times 112.993 \dots \times 158.472 \dots \times \sin 100$ (A1)

$\frac{1}{2} \times 210 \dots \times 158.472 \dots \times sin (their part (a))$ $\frac{1}{2} \times 210 \dots \times 158.472 \dots \times \sin (their part (a))$ (A1)

THEN

$8820 m^{2} (8817.18 \dots)$ $8820 m^{2} (8817.18 \dots)$ A1

METHOD 3

${CE}^{2} = 210^{2} + 112.993 \dots^{2} - (2 \times 210 \times 112.993 \dots \times cos (180 - 100 - their part (a)))$ ${CE}^{2} = 210^{2} + 112.993 \dots^{2} - (2 \times 210 \times 112.993 \dots \times \cos (180 - 100 - their part (a)))$ (M1)

$(CE =) 158.472 \dots$ $(CE =) 158.472 \dots$ (A1)

substituted area of triangle formula (M1)

$\frac{1}{2} \times 112.993 \dots \times 158.472 \dots \times sin 100$ $\frac{1}{2} \times 112.993 \dots \times 158.472 \dots \times \sin 100$ (A1)

$8820 m^{2} (8817.18 \dots)$ $8820 m^{2} (8817.18 \dots)$ A1

[5 marks]

$1005 - 210$ $1005 - 210$ OR $795$ $795$ (A1)

equating answer to part (c) to area of a triangle formula (M1)

$8817.18 \dots = \frac{1}{2} \times DF \times (1005 - 210) \times sin 48.002 \dots °$ $8817.18 \dots = \frac{1}{2} \times DF \times (1005 - 210) \times \sin 48.002 \dots °$ (A1)

$(DF=) 29.8 m (29.8473 \dots)$ $(DF=) 29.8 m (29.8473 \dots)$ A1

[4 marks]

Examiners report

[N/A]

Syllabus sections

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

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