Question

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

Answer 1

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

To find: $\angle BAC$

$\angle D+\angle B={180}^{\circ }\text{(Opposite angles of a cyclic quadilateral )}$

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

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

$\angle ACB={90}^{\circ }\text{(Angle in a semicircle)}$

$\text{In}\Delta ABC,$

$\angle BAC+{50}^{\circ }+{90}^{\circ }={180}^{\circ }\text{(Since sum of angles of a triangle is}{180}^{\circ })$

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