Question

In the given figure, seg AB and seg AD are the chords of the circle. C is the point on the tangent of the circle at point A.
If m(arc APB) = 80$°$ and $\angle$BAD = 30$°$ then find
(i) $\angle$BAC
(ii) m(arc BQD)

Answer 1

Construction: Join OD, OB and OA.

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

$$\begin{gathered} \mathrm{m}(\mathrm{arc}\mathrm{BQD})=2\angle \mathrm{BAD}=2\times 30°=60° \\ \mathrm{m}(\mathrm{arc}\mathrm{APB})=80° \\ \mathrm{Now},\angle \mathrm{AOD}=\mathrm{m}(\mathrm{arc}\mathrm{BQD})+\mathrm{m}(\mathrm{arc}\mathrm{APB})=60°+80°=140° \\ \mathrm{Also}\mathit{,}\mathit{}\angle \mathrm{ODA}=\angle \mathrm{OAD}(\mathrm{Angles}\mathrm{opposite}\mathrm{to}\mathrm{equal}\mathrm{sides}) \\ \mathrm{In}△\mathrm{AOD},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{AOD}+\angle \mathrm{ODA}+\angle \mathrm{OAD}=180°(\mathrm{Angle}\mathrm{sum}\mathrm{property}) \\ \Rightarrow 2\angle \mathrm{OAD}=180°-140° \\ \Rightarrow \angle \mathrm{OAD}=20°...\left(\mathrm{i}\right) \\ \angle \mathrm{OAC}=90°(\mathrm{Radius}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{tangent}).....\left(\mathrm{ii}\right) \\ \therefore \angle \mathrm{BAC}=\angle \mathrm{OAC}-\angle \mathrm{OAD}-\angle \mathrm{BAD}=90°-20°-30°=40°\left(\mathrm{from}\right(\mathrm{i})\mathrm{and}(\mathrm{ii}\left)\right) \end{gathered}$$

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

$\therefore \mathrm{m}(\mathrm{arc}\mathrm{BQD})=2\angle \mathrm{BAD}=2\times 30°=60°$