Question

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

• A. ${50}^{\circ }$
• B. ${25}^{\circ }$
• C. ${60}^{\circ }$
• D. ${40}^{\circ }$

Answer 1

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

Given: ABCD is a cyclic quadrilateral whose side AB is the diameter of the circle and $\angle ADC={130}^{\circ }$.

To find $\angle BAC:$

Since opposite angles of a cyclic quadilateral are supplementary, we have
$\angle D+\angle B={180}^{\circ }.$

i.e., ${130}^{\circ }+\angle B={180}^{\circ }$

$⟹\angle B={180}^{\circ }-{130}^{\circ }={50}^{\circ }$

Since an angle in a semicircle is a right angle, we have
$\angle ACB={90}^{\circ }.$

$\text{In}\Delta ABC,$

$\angle BAC+{50}^{\circ }+{90}^{\circ }={180}^{\circ }.\text{(Angle sum property})$

$⟹\angle BAC={180}^{\circ }-{90}^{\circ }-{50}^{\circ }={40}^{\circ }$