Question

$ABCD$ is cyclic quadrilateral such that $AB$ is a diameter of the circle circumscribing it and $\angle ADC={140}^{\circ },$ then $\angle BAC$ is equal to

(A) ${80}^{\circ }$
(B) ${50}^{\circ }$
(C) ${40}^{\circ }$
(D) ${30}^{\circ }$

Answer 1

The correct option is (B): ${50}^{\circ }$

Given $ABCD$ is a cyclic quadrilateral and $\angle ADC={140}^{\circ }$

We know that, sum of the opposite angles in a cyclic quadrilateral is ${180}^{\circ }$
$\angle ADC+\angle ABC={180}^{\circ }$
$\Rightarrow {140}^{\circ }+\angle ABC={180}^{\circ }$
$\Rightarrow \angle ABC={180}^{\circ }-{140}^{\circ }$
$\therefore \angle ABC={40}^{\circ }$
Since $\angle ACB$ is an angle which lies in a semi-circle.
$\therefore \angle ACB={90}^{\circ }$ [$\because$ Angle in a semicircle is a right angle.]

$In\Delta ABC,$ we have

$\angle BAC+\angle ACB+\angle ABC={180}^{\circ }$ [by angle sum property of a triangle]

$\Rightarrow \angle BAC+{90}^{\circ }+{40}^{\circ }={180}^{\circ }$

$\Rightarrow \angle BAC={180}^{\circ }-{130}^{\circ }={50}^{\circ }$