WAEC 2012 · Paper 2 · Q10

(a)
If $3^x + 3^{x + 1} = 36$ , find $x$ .
(b)
The bearing of $P$ from $X$ , 10 km away, is $025^\circ$ . Another point $Q$ is 6 km from $X$ on a bearing of $162^\circ$ . Calculate the: (i) distance $PQ$ ; (ii) bearing of $P$ from $Q$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

$3^{x + 1} = 3 \times 3^x$ , so $3^x + 3 \times 3^x = 36$ , which is $4 \times 3^x = 36$ .
So $3^x = 9 = 3^2$ and $x = 2$ .

(b)

Draw north at $X$ .
$P$ is 10 km away on $025^\circ$ and $Q$ is 6 km away on $162^\circ$ .
The angle between the two lines at $X$ is $\angle PXQ = 162^\circ - 25^\circ$
$= 137^\circ$ .

(i)

Cosine rule: $|PQ|^2 = 10^2 + 6^2 - 2(10)(6)\cos 137^\circ$
$= 136 + 120(0.7314)$
$\approx 223.76$ .
So $|PQ| \approx 14.96\text{ km}$ .

(ii)

Sine rule for the angle at $Q$ : $\sin\angle PQX = \frac{10\sin 137^\circ}{14.96}$
$\approx \frac{6.820}{14.96}$
$\approx 0.4559$ , so $\angle PQX \approx 27.1^\circ$ .
At $Q$ , the direction to $X$ is the back bearing of $162^\circ$ , which is $342^\circ$ .
$P$ is $27.1^\circ$ further round clockwise: $342^\circ + 27.1^\circ = 369.1^\circ$ , which is $009^\circ$ (take away $360^\circ$ ).

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