Question 134

PA and PB are two tangents to a circle with centre 0. from a point P outside the circle. A and B are points on the circle. If $$\angle APB=70^\circ$$. then $$\angle OAB$$ is equal to:

Solution

$$\angle APB=70°$$

Then $$\angle AOB=110°$$

But triangle AOB is isosceles triangle as

$$AO=BO.$$

So,$$\angle OAB=\angle OBA.$$

So,$$2\angle OAB + \angle AOB=180°$$

or,$$\angle OAB=70/2=35°.$$

B is correct choice.

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