WAEC 2020 · Paper 2 · Q11

(a)
In the diagram, $O$ is the centre of circle $ABCD$ such that $\angle AOB = 98^\circ$ , $\angle DBA = 68^\circ$ and $\angle BDC = 47^\circ$ . Find: (i) $\angle CBD$ ; (ii) $\angle DCB$ .
Separate values with commas, e.g. 3, −2
(b)
Given that $10\cos(x + 17^\circ) - 2 = 0$ , $0^\circ \le x \le 90^\circ$ , calculate, correct to the nearest degree, the value of $x$ .
(c)
The sum of the interior angles of a regular polygon with $n$ sides is $(120n)^\circ$ . Find the value of $n$ .

Worked solution (try it first)

$\angle ADB$ stands on the same arc $AB$ as the angle $AOB$ at the centre, so $\angle ADB = \frac12 \times 98^\circ$
$= 49^\circ$ .
Then $\angle ADC = \angle ADB + \angle BDC$
$= 49^\circ + 47^\circ$
$= 96^\circ$ .
$ABCD$ is cyclic, so $\angle ABC = 180^\circ - 96^\circ$
$= 84^\circ$ (opposite angles).
So $\angle CBD = \angle ABC - \angle ABD$
$= 84^\circ - 68^\circ$
$= 16^\circ$ .

In triangle $BCD$ : $\angle DCB = 180^\circ - 47^\circ - 16^\circ$
$= 117^\circ$ .

$10\cos(x + 17^\circ) = 2$ , so $\cos(x + 17^\circ) = 0.2$ .
Then $x + 17^\circ = \cos^{-1} 0.2$
$\approx 78.46^\circ$ , and $x \approx 61.46^\circ \approx 61^\circ$ .

The interior angles of an $n$ -sided polygon add up to $(n - 2) \times 180^\circ$ .
So $(n - 2) \times 180 = 120n$ .
Then $180n - 360 = 120n$ , $60n = 360$ and $n = 6$ .