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Date	May 2022	Marks available	5	Reference code	22M.2.SL.TZ2.7
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 2
Command term	Find	Question number	7	Adapted from	N/A

Question

A farmer is placing posts at points $A$ , $B$ , and $C$ in the ground to mark the boundaries of a triangular piece of land on his property.

From point $A$ , he walks due west $230$ metres to point $B$ .
From point $B$ , he walks $175$ metres on a bearing of $063 °$ to reach point $C$ .

This is shown in the following diagram.

The farmer wants to divide the piece of land into two sections. He will put a post at point $D$ , which is between $A$ and $C$ . He wants the boundary $BD$ to divide the piece of land such that the sections have equal area. This is shown in the following diagram.

Find the distance from point $A$ to point $C$ .

[4]

Find the area of this piece of land.

[2]

Find $CÂB$ .

[3]

Find the distance from point $B$ to point $D$ .

[5]

Markscheme

$A \hat{B} C = 27 °$ (A1)

attempt to substitute into cosine rule (M1)

$175^{2} + 230^{2} - 2 (175) (230) \cos 27 °$ (A1)

$108.62308 \dots$

$AC = 109 (m)$ A1

[4 marks]

correct substitution into area formula (A1)

$\frac{1}{2} \times 175 \times 230 \times \sin 27 °$

$9136.55 \dots$

area $= 9140 (m^{2})$ A1

[2 marks]

attempt to substitute into sine rule or cosine rule (M1)

$\frac{\sin 27 °}{108.623 \dots} = \frac{\sin \hat{A}}{175}$ OR $\cos A = \frac{{(108.623 \dots)}^{2} + 230^{2} - 175^{2}}{2 \times 108.623 \dots \times 230}$ (A1)

$47.0049 \dots$

$CÂB = 47.0 °$ A1

[3 marks]

METHOD 1

recognizing that for areas to be equal, $AD = DC$ (M1)

$AD = \frac{1}{2} AC = 54.3115 \dots$ A1

attempt to substitute into cosine rule to find $BD$ (M1)

correct substitution into cosine rule (A1)

${BD}^{2} = 230^{2} + 54 . 3115^{2} - 2 (230) (54.3115) \cos 47.0049 °$

$BD = 197.009 \dots$

$BD = 197 (m)$ A1

METHOD 2

correct expressions for areas of triangle $BDA$ and triangle $BCD$ using $BD$ A1

$\frac{1}{2} \times BD \times 230 \times \sin x °$ and $\frac{1}{2} \times BD \times 175 \times \sin (27 - x) °$ OR

$\frac{1}{2} \times BD \times 230 \times \sin (27 - x) °$ and $\frac{1}{2} \times BD \times 175 \times \sin x °$

correct equation in terms of $x$ (A1)

$175 \sin (27 - x) = 230 \sin x$ or $175 \sin x = 230 \sin (27 - x)$

$x = 11.6326 \dots$ or $x = 15.3673 \dots$ (A1)

substituting their value of $x$ into equation to solve for $BD$ (M1)

$\frac{1}{2} \times BD \times 230 \times \sin 11.6326 \dots = \frac{1}{2} \times BD \times 175 \times \sin 15.3673 \dots$ or

$\frac{1}{2} \times BD \times 230 \times \sin 11.6326 \dots = \frac{1}{2} \times 9136.55 \dots$

$BD = 197 (m)$ A1

[5 marks]

Examiners report

Students performed well on parts (a)-(c), correctly applying the cosine rule, the sine formula for area and the sine rule. Part (d) proved challenging. A common error was to falsely assume that segment BD bisected angle ABC.

A significant number of candidates did not have their calculator in degree mode or started in radians and changed to degrees part way through but used answers they had obtained when they were in radian mode. They got answers which were clearly impossible from the diagram, but most did not notice this.

Accuracy was a great problem throughout this question: premature rounding, incorrect rounding, or quoting more figures for the answer than they had used in the calculation.

[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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Find the size of angle ${\text{B}}\mathop {\text{A}}\limits^ \wedge {\text{C}}$ .
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Find the size of angle ${\text{C}}\mathop {\text{A}}\limits^ \wedge {\text{D}}$ .
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Find the size of angle ${\text{A}}\mathop {\text{C}}\limits^ \wedge {\text{D}}$ .
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16N.2.SL.TZ0.T_5c:
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16N.2.SL.TZ0.T_5e:
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Topic 3— Geometry and trigonometry

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