WAEC 2022 · Paper 2 · Q4

Circle geometry

In the diagram, $BCDE$ is a circle with centre $A$ . $\angle BCD = (2x + 40)^\circ$ , $\angle BAD = (5x - 35)^\circ$ , $\angle BED = (2y + 10)^\circ$ and $\angle ADC = 40^\circ$ . Find:

(a)
the values of $x$ and $y$ ;
Separate values with commas, e.g. 3, −2
(b)
$\angle ABC$ .

Worked solution (try it first)

(a)

$BCDE$ is a cyclic quadrilateral, so opposite angles add up to $180^\circ$ : $(2x + 40) + (2y + 10) = 180$ , which gives $x + y = 65$ .
The angle at the centre is twice the angle at the circumference on the same arc $BD$ : $5x - 35 = 2(2y + 10)$ , which gives $5x - 4y = 55$ .
From the first equation, $x = 65 - y$ .
Substitute: $5(65 - y) - 4y = 55$ , so $325 - 9y = 55$ , $9y = 270$ and $y = 30$ .
Then $x = 35$ .

(b)

With $x = 35$ : $\angle BCD = 110^\circ$ and $\angle BAD = 140^\circ$ .
The angles of quadrilateral $ABCD$ add up to $360^\circ$ .
So $\angle ABC = 360^\circ - 140^\circ - 110^\circ - 40^\circ$
$= 70^\circ$ .

Report a problem with this question