WAEC 2009 · Paper 2 · Q7

(a)
A woman looking out from the window of a building at a height of 30 m observed that the angle of depression of the top of a flag pole was $44^\circ$ . If the foot of the pole is 25 m from the foot of the building and on the same horizontal ground, find, correct to the nearest whole number, the: (i) angle of depression of the foot of the pole from the woman; (ii) height of the flag pole.
Separate values with commas, e.g. 3, −2
(b)
In the diagram, $O$ is the centre of the circle, $\angle OQR = 32^\circ$ and $\angle TPQ = 15^\circ$ . Calculate: (i) $\angle QPR$ ; (ii) $\angle TQO$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

The angle of depression $x$ of the foot equals the angle of elevation of $W$ from $O$ : $\tan x = \frac{30}{25} = 1.2$ .
So $x = \tan^{-1}(1.2) = 50.2^\circ$ , which is $50^\circ$ to the nearest degree.

$|OQ| = |OR|$ (radii), so $\angle ORQ = 32^\circ$ and $\angle QOR = 180^\circ - 2(32^\circ)$
$= 116^\circ$ .
The angle at the centre is twice the angle at the circumference: $\angle QPR = \frac{116^\circ}{2}$
$= 58^\circ$ .

$\angle TPR = \angle TPQ + \angle QPR$
$= 15^\circ + 58^\circ$
$= 73^\circ$ .
$QTPR$ is a cyclic quadrilateral, so opposite angles add to $180^\circ$ : $\angle TQR = 180^\circ - 73^\circ$
$= 107^\circ$ .
$\angle TQO = \angle TQR - \angle OQR$
$= 107^\circ - 32^\circ$
$= 75^\circ$ .