Question

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

• A. ${40}^{\circ }$
• B. ${50}^{\circ }$
• C. ${60}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

The correct option is A ${40}^{\circ }$

Construction: Join A and C. This forms a right triangle ACB [ angle in a semicircle is ${90}^{\circ }$]

From the figure, we can say that $\angle \text{B}={50}^{\circ }$ [opposite angles in a cyclic quadrilateral are supplementary]

Consider triangle ACB.

$\angle \text{ACB}={90}^{\circ }$ [ angle in a semicircle is ${90}^{\circ }$]

$\angle \text{ABC}={50}^{\circ }$

$⟹\angle \text{BAC}={40}^{\circ }$ [angle sum property of a triangle].