Question

Question 8
In figure, if PA and PB are tangents to the circle with centre O such that $\angle APB={50}^{\circ }$, then $\angle$ OAB is equal to
(A) 25°
(B) 30°
(C) 40°
(D) 50°

Answer 1

Given. PA and PB are tangent lines.
PA = PB
[ Since , the length of tangents drawn from an external point to a circle is equal ]
$\Rightarrow \angle PBA=\angle PAB=\theta \left[say\right]In\Delta PAB,\angle P+\angle A+\angle B={180}^{\circ }[Since,sumofanglesofatriangle={180}^{\circ }]\Rightarrow {50}^{\circ }+\theta +\theta ={180}^{\circ }\Rightarrow 2\theta ={180}^{\circ }-{50}^{\circ }={130}^{\circ }\Rightarrow \theta ={65}^{\circ }Also,OA\perp PA$
[Since , tangents at any point of a circle is perpendicular to the radius through the point of contact]
$\therefore \angle PAO={90}^{\circ }\Rightarrow \angle PAB+\angle BAO={90}^{\circ }\Rightarrow {65}^{\circ }+\angle BAO={90}^{\circ }\Rightarrow \angle BAO={90}^{\circ }–{65}^{\circ }={25}^{\circ }$