WAEC 2025 · Paper 1 · Q34✱✱

In the diagram, $O$ is the centre of the circle, $AOD$ is produced to $C$ and $|BD| = |DC|$ . If $\angle DCB = 35^\circ$ , find $\angle BAO$ .

Worked solution (try it first)

$|BD| = |DC|$ , so triangle $BDC$ is isosceles and $\angle DBC = \angle DCB = 35^\circ$ .
$\angle BDA$ is an exterior angle of triangle $BDC$ : $\angle BDA = 35^\circ + 35^\circ$
$= 70^\circ$ .
$AD$ is a diameter, so $\angle ABD = 90^\circ$ .
In triangle $ABD$ , $\angle BAO = 180^\circ - 90^\circ - 70^\circ$
$= 20^\circ$ , option C.