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Date	May 2019	Marks available	6	Reference code	19M.1.AHL.TZ1.H_4
Level	Additional Higher Level	Paper	Paper 1	Time zone	Time zone 1
Command term	Find	Question number	H_4	Adapted from	N/A

Question

The lengths of two of the sides in a triangle are 4 cm and 5 cm. Let θ be the angle between the two given sides. The triangle has an area of $\frac{{5\sqrt {15} }}{2}$ cm².

Show that ${\text{sin}}\,\theta = \frac{{\sqrt {15} }}{4}$ .

[1]

a.

Find the two possible values for the length of the third side.

[6]

b.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

EITHER

$\frac{{5\sqrt {15} }}{2} = \frac{1}{2} \times 4 \times 5\,{\text{sin}}\,\theta$ A1

OR

height of triangle is $\frac{{5\sqrt {15} }}{4}$ if using 4 as the base or ${\sqrt {15} }$ if using 5 as the base A1

THEN

${\text{sin}}\,\theta = \frac{{\sqrt {15} }}{4}$ AG

[1 mark]

a.

let the third side be $x$

${x^2} = {4^2} + {5^2} - 2 \times 4 \times 5 \times {\text{cos}}\,\theta$ M1

valid attempt to find ${\text{cos}}\,\theta$ (M1)

Note: Do not accept writing ${\text{cos}}\left( {{\text{arcsin}}\left( {\frac{{\sqrt {15} }}{4}} \right)} \right)$ as a valid method.

${\text{cos}}\,\theta = \pm \sqrt {1 - \frac{{15}}{{16}}}$

$= \frac{1}{4}{\text{,}}\,\, - \frac{1}{4}$ A1A1

${x^2} = 16 + 25 - 2 \times 4 \times 5 \times \pm \frac{1}{4}$

$x = \sqrt {31}$ or $\sqrt {51}$ A1A1

[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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