Question

In the given figure, $ABCD$ is a cyclic quadrilateral whose side $AB$ is a diameter of the circle passing through $A,B,C$ and $D$. If $\angle$ADC$={130}^o$, find $\angle$BAC.

• A. ${45}^o$
• B. ${58}^o$
• C. ${60}^o$
• D. ${40}^o$

Answer 1

The correct option is D ${40}^o$

$\angle ACB={90}^o$ (angle subtended by the diameter of circle)

and $\angle ABC$ is the supplementary angle of $\angle ADC$...(opposite angles of cyclic quadrilateral)

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

Now, by angle sum property,

$\angle BAC=180-(\angle ACB+\angle ABC)=180-50-90={40}^o$.

Hence, option $D$ is correct.