Question 114

PA and PB are two tangents to a circle with centre O, from a point P outside the circle. A and B are points on the circle. If $$\angle PAB = 47^\circ$$, then $$\angle OAB$$ is equal to:

Solution

We know that tangents drawn from a same point outside circle are of equal length.

So, triangle PAB is an isosceles triangle.

So,$$\angle PAB=\angle PBA=47°$$

$$\angle APB=180°-94°=86°$$

again, triangle OAB is isosceles triangle.

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

and $$\angle AOB + \angle APB =180°$$

So,$$\angle AOB= 94°$$

So,$$\angle OAB=(180-94)°/2=43°$$

So, Option A is correct choice.

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