NECO 2023 · Paper 1 · Q41

In the diagram, $O$ is the centre of the circle, $AB$ is a diameter, $|DB| = |BC|$ and $\angle ABC = 54^\circ$ . Find $\angle ACD$ .

Worked solution (try it first)

$DB = BC$ , so triangle $DBC$ is isosceles with apex $54^\circ$ at $B$ : $\angle BCD = \dfrac{180^\circ - 54^\circ}{2}$
$= 63^\circ$ .
$AB$ is a diameter, so $\angle ACB = 90^\circ$ (angle in a semicircle).
$\angle ACD$ is the rest of that right angle: $\angle ACD = 90^\circ - 63^\circ$
$= 27^\circ$ , option E.