Question

A quadrilateral $ABCD$ is inscribed in a circle such that $AB$ is a diameter and $\angle ADC={130}^o$, then $m\angle BAC=$?

• A. ${90}^o$
• B. ${50}^o$
• C. ${40}^o$
• D. ${30}^o$

Answer 1

Answer: (C)

${40}^o$

Given, $ABCD$ is a quadrilateral inscribed in a circle.

Also, $\angle ADC={130}^o$.

We have to find: $\angle BAC$.

Now join $AD$.

Then, $\angle ADC+\angle ABC={180}^o$ ....(The sum of the opposite angles of a cyclic quadrilateral $={180}^o$)

i.e. $\angle ABC={180}^o-\angle ADC={180}^o-{130}^o={50}^o$.

So, in $\Delta ABC$, we have

$\angle BAC={180}^o$ $-(\angle ACB+\angle ABC)$ ...(Since $AB$ is the diameter, the angle subtended by the diameter of a circle at the circumference $={90}^o$ i.e. $\angle BAC={90}^o$)

$={180}^o-({50}^o+{90}^o)={40}^o$.

Therefore, option $C$ is correct.