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Date	November 2019	Marks available	5	Reference code	19N.2.SL.TZ0.T_5
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Determine and Justify	Question number	T_5	Adapted from	N/A

Question

Haraya owns two triangular plots of land, $ABC$ and $ACD$ . The length of $AB$ is $30 m$ , $BC$ is $50 m$ and $AC$ is $70 m$ . The size of $D \hat{A} C$ is $55 °$ and $A \hat{D} C$ is $72 °$ .

The following diagram shows this information.

Haraya attaches a $20 m$ long rope to a vertical pole at point $B$ .

Find the length of $AD$ .

[4]

Find the size of $A \hat{B} C$ .

[3]

Calculate the area of the triangular plot of land $ABC$ .

[3]

Determine whether the rope can extend into the triangular plot of land, $ACD$ . Justify your answer.

[5]

Markscheme

$A \hat{C} D = 53 °$ (or equivalent) (A1)

Note: Award (A1) for $53 °$ (or equivalent) seen.

$\frac{AD}{\sin 53 °} = \frac{70}{\sin 72 °}$ (M1)(A1)

Note: Award (M1) for substitution into sine rule formula, (A1) for correct substitution.

$({AD}^{2} =) 60.2915 \dots^{2} + 70^{2} - 2 \times 70 \times 60.2915 \dots \times \cos 53$ (A1)(M1)(A1)

Note: Award (A1) for $53$ or $60.2915 . . .$ seen, (M1) for substitution into cosine rule formula, (A1) for correct substitution.

$(AD =) 58.8 (m) (58.7814 \dots)$ (A1)(G3)

[4 marks]

$(\cos A \hat{B} C) = \frac{30^{2} + 50^{2} - 70^{2}}{2 \times 30 \times 50}$ (M1)(A1)

Note: Award (M1) for substitution into cosine rule formula, (A1) for correct substitution.

$(A \hat{B} C =) 120 °$ (A1)(G2)

[3 marks]

Units are required in part (c)

$A = \frac{1}{2} \times 50 \times 30 \times \sin 120 °$ (M1)(A1)(ft)

Note: Award (M1) for substitution into the area formula, (A1)(ft) for correct substitution. Award (M0)(A0)(A0) for $\frac{1}{2} \times 50 \times 30$ .

$(A =) 650 m^{2} (649.519 \dots m^{2})$ (A1)(ft)(G2)

Note: Follow through from part (b).

[3 marks]

METHOD 1 (equating part (c) to expression for area of triangle ABC)

$649.519 \dots = \frac{1}{2} \times 70 \times h$ (M1)(A1)(ft)

Note: Award (M1) for correctly substituted area of triangle formula. Award (A1)(ft) for equating the area formula to their area found in part (c).

$(h =) 18.6 (m) (18.5576 \dots)$ (A1)(ft)

Note: Follow through from their part (c).

$20 > 18.5576 \dots$ (R1)(ft)

Note: Accept “the length of the rope is greater than the altitude of triangle $ABC$ ”.

the rope passes inside the triangular plot of land $ACD$ (A1)(ft)

Note: Follow through from their altitude. The final (A1) is contingent on (R1) being awarded.

METHOD 2 (finding $C \hat{A} B$ or $A \hat{C} B$ with sine rule and then trig ratio)

$\frac{\sin C \hat{A} B}{50} = \frac{\sin 120 °}{50} (C \hat{A} B = 38.2132 \dots °)$ (M1)

Note: Award (M1) for their correct substitution into sine rule formula to find $C \hat{A} B$ or $A \hat{C} B$ . Follow through from their part (b).

$(h =) 30 \times \sin (38.2132 \dots °)$ (M1)

Note: Award (M1) for correct substitution of their $C \hat{A} B$ or $A \hat{C} B$ into trig formula.

$(h =) 18.6 (m) (18.5576 \dots)$ (A1)(ft)

Note: Follow through from their part (b).

$20 > 18.5576 \dots$ (R1)(ft)

Note: Accept “the length of the rope is greater than the altitude of triangle $ABC$ ”.

the rope passes inside the triangular plot of land $ACD$ (A1)(ft)

Note: Follow through from their altitude. The final (A1) is contingent on (R1) being awarded.

METHOD 3 (finding $C \hat{A} B$ or $A \hat{C} B$ with with cosine rule and then trig ratio)

$\cos A \hat{C} B = \frac{50^{2} + 70^{2} - 30^{2}}{2 (50) (70)} (A \hat{C} B = 21.7867 \dots °)$ (M1)

Note: Award (M1) for for their correct substitution into cosine rule formula to find $C \hat{A} B$ or $A \hat{C} B$ .

$(h =) 50 \times \sin (21.7867 \dots °)$ (M1)

Note: Award (M1) for correct substitution of their $C \hat{A} B$ or $A \hat{C} B$ into trig formula.

$(h =) 18.6 (m) (18.5576 \dots)$ (A1)(ft)

$20 > 18.5576 \dots$ (R1)(ft)

Note: Accept “the length of the rope is greater than the altitude of triangle $ABC$ ”.

the rope passes inside the triangular plot of land $ACD$ (A1)(ft)

Note: Follow through from their altitude. The final (A1) is contingent on (R1) being awarded.

METHOD 4 (finding area of triangle with height $20$ , justifying the contradiction)

$A = \frac{1}{2} (70) (20) = 700 (m^{2})$ (M1)(A1)

Note: Award (M1) for correct substitution into area of a triangle formula for a triangle with height $20$ and base $70$ . Award (A1) for $700$ . Award (M0)(A0) for unsupported $700$ unless subsequent reasoning explains how the $700$ was found.

$700 > 649.519 \dots$ (R1)

if rope exactly touches the $AC$ then this triangle has an area greater than
$ABC$ and as the distance $AC$ is fixed the altitude must be less than $20$ (R1)

$\frac{1}{2} (70) (20) > \frac{1}{2} (70)$ (height perpendicular to $AC$ ) and therefore $20 >$ height perpendicular to $AC$ (R1)(ft)

Note: Award (R1) for an explanation that recognizes the actual triangle $ABC$ and this new triangle have the same base $(70)$ and hence the height of triangle $ABC$ is less than $20$ .

therefore, the rope passes inside the triangular plot of land $ACD$ (A1)(ft)

Note: Other methods, besides those listed here, may be possible. These methods can be summarized in two broad groups: the first is to find the altitude of the triangle, and compare it to $20$ , and the second is to create an artificial triangle with an altitude of $20$ and explain why this triangle is not $ABC$ by relating to area and the given lengths of the sides.

[5 marks]

Examiners report

[N/A]

Syllabus sections

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

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