Question

Chords AB and CD of a circle intersect in point Q in the interior of a circle as shown in the given figure.
If m (arc AD) = 25$°$ and m (arc BC) = 31$°$, then find $\angle$BQC

Answer 1

Construction: Join OD, OB, OA, BD and OC.

We know that the angle subtended by the arc at the centre is twice the angle subtended at any part of the circle.

$\angle$DBA = $\frac{1}{2}\mathrm{m}\left(\mathrm{arcAD}\right)=\frac{1}{2}\times 25°=12.5°$

$\angle$BDC = $\frac{1}{2}\mathrm{m}\left(\mathrm{arcBC}\right)=\frac{1}{2}\times 31°=15.5°$

$$\begin{gathered} \mathrm{In}△\mathrm{DBQ},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{BQC}=\angle \mathrm{DBA}+\angle \mathrm{BDC}(\mathrm{Exterior}\mathrm{angle}\mathrm{property}) \\ =12.5°+15.5° \\ =28° \end{gathered}$$