WAEC 2008 · Paper 2 · Q9

(a)
The triangle $ABC$ has sides $|AB| = 17\text{ m}$ , $|BC| = 12\text{ m}$ and $|AC| = 10\text{ m}$ . Calculate the: (i) largest angle of the triangle; (ii) area of the triangle.
Separate values with commas, e.g. 3, −2
(b)
From a point $T$ on a horizontal ground, the angle of elevation of the top $R$ of a tower $RS$ , $38\text{ m}$ high, is $63^\circ$ . Calculate, correct to the nearest metre, the distance between $T$ and $S$ .

Worked solution (try it first)

(a)(i)

The largest angle is opposite the longest side, $AB = 17$ , so it is angle $C$ .
Use the cosine rule: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
$= \frac{12^2 + 10^2 - 17^2}{2 \times 12 \times 10}$
$= \frac{-45}{240}$
$= -0.1875$ .
The cosine is negative, so $C$ is obtuse: $C = 180^\circ - 79.19^\circ$
$= 100.81^\circ$ .
The largest angle is $100.8^\circ$ .

(ii)

Area $= \frac12 ab\sin C$
$= \frac12 \times 12 \times 10 \times \sin 100.81^\circ$ .
$\sin 100.81^\circ = 0.9823$ , so the area is $60 \times 0.9823 = 58.94\text{ m}^2$ .

(b)

In the right-angled triangle $RST$ , $RS = 38$ is opposite the $63^\circ$ angle and $TS$ is adjacent.
$\tan 63^\circ = \frac{38}{TS}$ , so $TS = \frac{38}{\tan 63^\circ}$
$= \frac{38}{1.9626}$
$= 19.36$ .
The distance $TS$ is $19\text{ m}$ to the nearest metre.

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