Question

In the figure, $\Delta$APB is formed by three tangents to the circle with centre O. If $\angle APB={40}^o$ , then the measure of $\angle BOA$ is

• A. ${50}^o$
• B. ${55}^o$
• C. ${60}^o$
• D. ${70}^o$

Answer 1

The correct option is D ${70}^o$

From the figure, Join $OP$ and name the point of intersection of $OP$ and $BA$ as $S$.

Now, $PR=PT$

$\therefore OP$ is an angle bisector of $\angle RPA$

$\therefore \angle OPR={20}^o=\angle OPT$

As, $\angle ORP={90}^o=\angle OTP$ ........... (Radius and tangents are perpendicular to each other)

In $△ORP$,

$\angle ORP+\angle OPR+\angle POR={180}^o$

$\therefore \angle POR={70}^o$

Also, $\angle POT=\angle POT={70}^o$ ........... (Equal tangents subtend equal angles)

Similarly, $\angle ROB=\angle BOS$ and $\angle TOA=\angle AOS$

$\therefore$ $(\angle ROB+\angle BOS)+(\angle TOA+\angle AOS)=\angle POR+\angle POT={70}^o+{70}^o={140}^o$

$⟹2\angle BOS+2\angle AOS={140}^o$

$⟹\angle BOS+\angle AOS={70}^o$

$\therefore \angle BOA={70}^o$.

Hence, option D is correct.