Question

PA and PB are two tangents drawn from an external point P to a circle with centre O. If ∠APB = 80° then ∠POA = ______.

Answer 1

PA and PB are two tangents drawn from an external point P to a circle with centre O. If ∠APB = 80° then ∠POA = 50°.

Explanation:

\[ \begin{array}{r l} \textbf{Given:} & \text{Tangents } PA \text{ and } PB \text{ drawn from an external point } P \text{ to a circle with centre } O, \text{ with } \angle APB = 80^\circ. \\[4pt] \textbf{To Find:} & \text{The measure of } \angle POA. \\[4pt] \textbf{Solution:} & \text{The radius through the point of contact is perpendicular to the tangent, hence } \angle OAP = 90^\circ. \\[4pt] & \text{The line joining the centre to an external point bisects the angle between the two tangents, hence } \angle APO = \tfrac{1}{2}\angle APB. \\[4pt] & \text{In the triangle } OAP, \text{ by the angle sum property of a triangle:} \\[4pt] & \begin{aligned} \angle APO &= \frac{1}{2} \times 80^\circ \\[4pt] &= 40^\circ \\[4pt] \angle POA &= 180^\circ - \angle OAP - \angle APO \\[4pt] &= 180^\circ - 90^\circ - 40^\circ \\[4pt] &= 50^\circ \\[4pt] &= 50.00^\circ \end{aligned} \\[4pt] \textbf{Answer:} & \angle POA = 50^\circ = 50.00^\circ. \end{array} \]