Question

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

• A. ${90}^{\circ }$
• B. ${20}^{\circ }$
• C. ${60}^{\circ }$
• D. ${50}^{\circ }$

Answer 1

The correct option is C ${60}^{\circ }$

Join OC
Given: $\angle$ AOB = ${140}^{\circ }$
In $△$ OAB,
OA = OB [$\because$ radius of circle]
By Theorem- Angles opposite to equal sides of a triangle are equal.
$\therefore$ $\angle$ OAB = $\angle$ OBA ---- (i)

By Theorem- Sum of angles of a triangle = ${180}^{\circ }$
$\therefore$ $\angle$ OAB + $\angle$ OBA + $\angle$ AOB = ${180}^{\circ }$
$\Rightarrow$ 2$\angle$ OAB + ${140}^{\circ }$ = ${180}^{\circ }$ [from (i)]

$\Rightarrow$ $\angle$ OAB = $\frac{({180}^{\circ }-{140}^{\circ })}{2}$
$\Rightarrow$ $\angle$ OAB = ${20}^{\circ }$

By Theorem- If two tangents PA and PC are drawn to a circle with centre O from an external point P, then $\angle$ APC = 2$\angle$ OAC = 2$\angle$ OCA

$\therefore$ $\angle$ OAC = $\frac{1}{2}$ $\angle$ APC
Given: $\angle$ APC = ${80}^{\circ }$
$\Rightarrow$ $\angle$ OAC = $\frac{{80}^{\circ }}{2}$
$\therefore$ $\angle$ OAC = ${40}^{\circ }$

Now,
The required $\angle$BAC = $\angle$OAB + $\angle$OAC
$\therefore$ $\angle$BAC = ${20}^{\circ }$ + ${40}^{\circ }$
$\therefore$ $\angle$BAC = ${60}^{\circ }$