JAMB 2004 · UME · Q33

Sine & cosine rules

In the triangle shown, the base angles are $45^\circ$ and $60^\circ$ , the side opposite the $60^\circ$ angle is 15 cm and the side opposite the $45^\circ$ angle is $x$ . Find $x$ .

Worked solution (try it first)

Sine rule: $x$ faces the $45^\circ$ angle and 15 cm faces the $60^\circ$ angle, so $\dfrac{x}{\sin45^\circ} = \dfrac{15}{\sin60^\circ}$ .
So $x = \dfrac{15 \times \frac{\sqrt2}{2}}{\frac{\sqrt3}{2}}$
$= \dfrac{15\sqrt2}{\sqrt3}$ .
Rationalise: multiply top and bottom by $\sqrt3$ to get $\dfrac{15\sqrt6}{3} = 5\sqrt6$ cm, option C.

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