Question

ABCD is a cyclic quadrilateral whose side AB is a diameter of the circle. If $\angle ADC={130}^{\circ }$, find $\angle BAC$.

• A. $(\frac{120}{2}{)}^{\circ }$
• B. $(\frac{120}{3}{)}^{\circ }$
• C. ${60}^{\circ }$
• D. ${40}^{\circ }$

Answer 1

Answer: (B), (D)

The correct options are
B

$(\frac{120}{3}{)}^{\circ }$

D

${40}^{\circ }$

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


$\angle D+\angle B={180}^{\circ }$ [Sum of opposite angles of a cyclic quadilateral is 180${}^{\circ }$]

$\Rightarrow {130}^{\circ }+\angle B={180}^{\circ }$

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

$\angle ACB={90}^{\circ }$ [Angle subtended by diameter on the circle is ${90}^{\circ }$]

In $\Delta ABC,$

$\angle BAC+\angle ACB+\angle CBA={180}^{\circ }$ [Sum of angles of a triangle is ${180}^{\circ }$]
$\angle BAC+{50}^{\circ }+{90}^{\circ }={180}^{\circ }$

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