Question

If tangents PA and PB from a point P to a circle with center O are inclined to each other at an angle of ${80}^{\circ }$ then find the measure of $\angle POA$

• A. ${50}^{\circ }$
• B. ${60}^{\circ }$
• C. ${70}^{\circ }$
• D. ${80}^{\circ }$

Answer 1

The correct option is A

${50}^{\circ }$

Given that,
$\angle$ APB = ${80}^{\circ }$

OA = OB (Radius of the circle)
PA = PB (Length of tangents from an external point to the circle are equal in length)
and OP is the common base

So, Δ AOP is congruent to $△$BOP SSS Postulate.

Therefore $\angle$ AOP = $\angle$ BOP (by CPCT)

We know that
$\angle$ AOB + $\angle$ OBP + $\angle$BPA + $\angle$PAO = ${360}^{\circ }$

Therefore $\angle$AOB + ${90}^{\circ }$ + ${80}^{\circ }$ + ${90}^{\circ }$ = ${360}^{\circ }$

$\angle$AOB = ${100}^{\circ }$ = 2$\angle$POA ($\angle$AOB = $\angle$POA + $\angle$POB and $\angle$POA = $\angle$POB)

Therefore, $\angle$POA = ${50}^{\circ }$