WAEC 2020 · Paper 2 · Q3

(a)
In the diagram, $O$ is the centre of the circle $ABCDE$ , $BE$ and $AD$ are diameters, $|BC| = |CD|$ and $\angle BCD = 108^\circ$ . Find $\angle CDE$ .
(b)
Given that $\tan x = \sqrt3$ , $0^\circ \le x \le 90^\circ$ , evaluate $\dfrac{(\cos x)^2 - \sin x}{(\sin x)^2 + \cos x}$ .

Worked solution (try it first)

(a)

In triangle $BCD$ , $|BC| = |CD|$ , so the base angles are equal: $\angle CDB = \frac{180^\circ - 108^\circ}{2}$
$= 36^\circ$ .
$BE$ is a diameter, so $\angle BDE = 90^\circ$ (angle in a semicircle).
So $\angle CDE = \angle CDB + \angle BDE$
$= 36^\circ + 90^\circ$
$= 126^\circ$ .

(b)

$\tan x = \sqrt3$ with $x$ acute, so $x = 60^\circ$ , $\sin x = \frac{\sqrt3}{2}$ and $\cos x = \frac12$ .
Top: $\left(\frac12\right)^2 - \frac{\sqrt3}{2} = \frac14 - \frac{\sqrt3}{2}$
$= \frac{1 - 2\sqrt3}{4}$ .
Bottom: $\left(\frac{\sqrt3}{2}\right)^2 + \frac12 = \frac34 + \frac12$
$= \frac54$ .
So the value is $\frac{1 - 2\sqrt3}{4} \times \frac45 = \frac{1 - 2\sqrt3}{5}$ .

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