Question

In the given figure, PA and PB are two tangents to the circle with centre O. If $\angle$ APB = ${60}^o$ then $\angle$ OAB is

(a) ${15}^o$ (b) ${30}^o$ (c) ${60}^o$ (d) ${90}^o$

Answer 1

In ΔAPB
angle APB=60°
AND
angle PAB= angle PBA
so
sum of angles in a triangle is 180
60°+angle PAB+angle PBA=180
2 angle PAB=180°-60°
PAB=120°/2=60°
And
angle OAP=90°
SO angle OAB=90°-60°=30°

option (B) 30° is correct answer