Question

PA and PB are two tangents drawn from an external point P to a circle with centre O. If ∠PBA = 65°, then ∠APB = ______.

Answer 1

PA and PB are two tangents drawn from an external point P to a circle with centre O. If ∠PBA = 65°, then ∠APB = 50°.

Explanation:

\[ \begin{array}{r l} \textbf{Given:} & \text{Tangents } PA \text{ and } PB \text{ drawn from an external point } P \text{ to a circle with centre } O, \text{ with } \angle PBA = 65^\circ. \\[4pt] \textbf{To Find:} & \text{The measure of } \angle APB. \\[4pt] \textbf{Solution:} & \text{The lengths of tangents drawn from an external point to a circle are equal, hence } PA = PB \text{ and the triangle } PAB \text{ is isosceles.} \\[4pt] & \text{In an isosceles triangle the angles opposite the equal sides are equal, hence } \angle PAB = \angle PBA. \\[4pt] & \text{In the triangle } PAB, \text{ by the angle sum property of a triangle:} \\[4pt] & \begin{aligned} \angle PAB &= \angle PBA \\[4pt] &= 65^\circ \\[4pt] \angle APB &= 180^\circ - \angle PAB - \angle PBA \\[4pt] &= 180^\circ - 65^\circ - 65^\circ \\[4pt] &= 50^\circ \\[4pt] &= 50.00^\circ \end{aligned} \\[4pt] \textbf{Answer:} & \angle APB = 50^\circ = 50.00^\circ. \end{array} \]