Question

In the given figure, O is the centre of the circumcircle ABC. Tangents at A and C intersect at P. Given angle $AOB={140}^{\circ }$ and angle $APC={80}^{\circ }$; find the angle BAC.

• A. ${60}^{\circ }$
• B. ${90}^{\circ }$
• C. ${50}^{\circ }$
• D. ${10}^{\circ }$

Answer 1

The correct option is A

${60}^{\circ }$

Join OC

$\therefore$ PA and PC are the tangents

$\therefore OA\perp$ and $OC\perp PC$

In quad APCO,

$\therefore \angle APC+\angle AOC={180}^{\circ }$

$\Rightarrow {80}^{\circ }+\angle AOC={180}^{\circ }$

$\therefore \angle AOC={180}^{\circ }-{80}^{\circ }={100}^{\circ }$

$\angle BOC={360}^{\circ }-(\angle AOB+\angle AOC)$

$={360}^{\circ }-({140}^{\circ }+{100}^{\circ })$

$={360}^{\circ }-{240}^{\circ }={120}^{\circ }$

Now arc BC subtends $\angle BOC$ at the circle

and $\angle BAC$ at the remaining part of the circle

$\therefore \angle BAC=\frac{1}{2}\angle BOC$

$=\frac{1}{2}\times {120}^{\circ }={60}^{\circ }$