NECO 2023 · Paper 1 · Q40

Circle geometry

In the figure, $O$ is the centre of the circle and $DB$ is a diameter. $AED$ and $ABC$ are straight lines, $\angle EAB = 34^\circ$ and $\angle EDB = 40^\circ$ . Calculate the value of $x$ .

Worked solution (try it first)

$DB$ is a diameter, so $\angle DEB = 90^\circ$ and, on the straight line $AED$ , $\angle AEB = 90^\circ$ .
Triangle $AEB$ gives $\angle ABE = 180^\circ - 90^\circ - 34^\circ$
$= 56^\circ$ .
Triangle $DEB$ gives $\angle EBD = 90^\circ - 40^\circ$
$= 50^\circ$ .
On the straight line $ABC$ , $\angle DBC = 180^\circ - 56^\circ - 50^\circ$
$= 74^\circ$ .
Angles in the same segment: $\angle ECB$ and $\angle EDB$ both stand on arc $EB$ , so $\angle ECB = 40^\circ$ .
In the triangle made by $B$ , $C$ and the crossing of $EC$ with $DB$ : $x = 180^\circ - 74^\circ - 40^\circ$
$= 66^\circ$ , option B.

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