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Date	November 2021	Marks available	6	Reference code	21N.1.AHL.TZ0.11
Level	Additional Higher Level	Paper	Paper 1	Time zone	Time zone 0
Command term	Find	Question number	11	Adapted from	N/A

Question

The following diagram shows a corner of a field bounded by two walls defined by lines $L_{1}$ and $L_{2}$ . The walls meet at a point $A$ , making an angle of $40 °$ .

Farmer Nate has $7 m$ of fencing to make a triangular enclosure for his sheep. One end of the fence is positioned at a point $B$ on $L_{2}$ , $10 m$ from $A$ . The other end of the fence will be positioned at some point $C$ on $L_{1}$ , as shown on the diagram.

He wants the enclosure to take up as little of the current field as possible.

Find the minimum possible area of the triangular enclosure $ABC$ .

Markscheme

METHOD 1

attempt to find $AC$ using cosine rule M1

$7^{2} = 10^{2} + {AC}^{2} - 2 \times 10 \times AC \times \cos 40 °$ (A1)

attempt to solve a quadratic equation (M1)

$AC = 4.888 \dots$ AND $10.432 \dots$ (A1)

Note: At least $AC = 4.888 \dots$ must be seen, or implied by subsequent working.

minimum area $= \frac{1}{2} \times 10 \times 4.888 \dots \times \sin (40 °)$ M1

Note: Do not award M1 if incorrect value for minimizing the area has been chosen.

$= 15.7 m^{2}$ A1

METHOD 2

attempt to find $A \hat{C} B$ using the sine Rule M1

$\frac{\sin C}{10} = \frac{\sin 40}{7}$ (A1)

$C = 66.674 \dots °$ OR $113.325 \dots °$ (A1)

EITHER

$B = 180 - 40 - 113.325 \dots$

$B = 26.675 \dots °$ (A1)

area $= \frac{1}{2} \times 10 \times 7 \times \sin (26.675 \dots °)$ M1

sine rule or cosine rule to find $AC = 4.888 \dots$ (A1)

minimum area $= \frac{1}{2} \times 10 \times 4.888 \dots \times \sin (40 °)$ M1

THEN

$= 15.7 m^{2}$ A1

Note: Award A0M1A0 if the wrong length $AC$ or the wrong angle $B$ selected but used correctly finding a value of $33.5 m^{2}$ for the area.

[6 marks]

Examiners report

As has often been the case in the past, trigonometry is a topic that is poorly understood and candidates are poorly prepared for. Approaches to this question required the use of the cosine or sine rules. Some candidates tried to use right-angled trigonometry instead. A minority of candidates used the cosine rule approach and were more likely to be successful, navigating the roots of the quadratic equation formed. When using the sine rule the method involved the ambiguous case as the required angle was obtuse. Few candidates realized this and this was the most common mistake. In a few instances, the word “minimum” led candidates to attempt an approach using calculus.

Syllabus sections

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

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