Question

If the tangents $PA$ and $PB$ from an external point $P$ to a circle with centre $O$ are inclined to each other at an angle of ${40}^{\circ }$ then $\angle POA=$

• A. ${140}^{\circ }$
• B. ${80}^{\circ }$
• C. ${70}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

Answer: (C)

${70}^{\circ }$

O is centre of circle.
PA and PB is external tangent.
So $\angle A$ and $\angle B$ = ${90}^{\circ }$
In quadrilateral OAPB
$\Rightarrow$ $\angle AOB$ + $\angle A$ +$\angle B$ + $\angle P$ = ${360}^{\circ }$
$\Rightarrow$ $\angle AOB$ = ${360}^{\circ }$ - ${220}^{\circ }$
$\therefore$ $\angle AOB$ = ${140}^{\circ }$

Now we can see that the $△AOP$ and $△POB$ are congruent to each other hence $\angle POA=\frac{1}{2}AOP=\frac{1}{2}\times {140}^o={70}^o$