Question

$ABCD$ is a 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 ${50}^{\circ }$

Since $ABCD$ is a cyclic quadrilateral,

the sum of the opposite angles will be $180$ degrees.

Thus, $\angle CBA+\angle ADC={180}^{\circ }$

$\Rightarrow$$\angle CBA={180}^{\circ }-\angle ADC$

$\Rightarrow$$\angle CBA={180}^{\circ }-{140}^{\circ }$

$\Rightarrow$$\angle CBA={40}^{\circ }$.

Also, angle in a semicircle in a right angle.

Therefore, $\angle ACB={90}^{\circ }$

Now, in $△ACB$,

$\angle CBA+\angle ACB+\angle BAC={180}^{\circ }$ ...[angle property of a triangle]

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

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

$\Rightarrow$$\angle BAC={50}^{\circ }$.

Hence, option $B$ is correct.