PA and PB are two tangents from a point P outside the circle with centre O. If A and are points on the circle such that $\angle APB = 100^\circ$, then $ \angle OAB$ is equal to:
Answer & explanation
Correct answer: option 4
A tangent to a circle forms a right angle with the circle's radius
= \(\angle\)OAP = \(\angle\)OBP = \({90}^\circ\)
The sum of all angles of a quadrilateral is \({360}^\circ\)
= \(\angle\)OAP + \(\angle\)OBP + \(\angle\)APB + \(\angle\)AOB = 360
= 90 + 90 + 100 + \(\angle\)AOB = 360
= \(\angle\)AOB = 360 - 280
= \(\angle\)AOB = \({80}^\circ\)
In triangle AOB
= \(\angle\)AOB + \(\angle\)OAB + \(\angle\)OBA = 180
= 80 + x + x = 180
= 2x = 100
= x = \({50}^\circ\)
Therefore, the value of \(\angle\)OAB is \({50}^\circ\).