WAEC 2022 · Paper 2 · Q12

Circle geometry

In the diagram, $W$ , $X$ , $Y$ , $Z$ are points on a circle centre $O$ . $\angle WXY = 111^\circ$ , $\angle OYX = 43^\circ$ and $|WZ| = |YZ|$ . Calculate:

(a)
$\angle OWX$ ;
(b)
$\angle OYZ$ .

Worked solution (try it first)

(a)

$WXYZ$ is a cyclic quadrilateral, so $\angle WZY = 180^\circ - 111^\circ$
$= 69^\circ$ .
The angle at the centre is twice the angle at the circumference on the same arc $WXY$ : $\angle WOY = 2 \times 69^\circ$
$= 138^\circ$ .
The angles of quadrilateral $OWXY$ add up to $360^\circ$ : $\angle OWX = 360^\circ - 138^\circ - 111^\circ - 43^\circ$
$= 68^\circ$ .

(b)

$|WZ| = |YZ|$ and $OW = OY$ , so $OZ$ is a line of symmetry and cuts $\angle WZY$ in half: $\angle OZY = \frac12 \times 69^\circ$
$= 34.5^\circ$ .
$OY = OZ$ (radii), so triangle $OYZ$ is isosceles and $\angle OYZ = \angle OZY = 34.5^\circ$ .

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