Question

Question 17
A quadrilateral ABCD is inscribed in a circle such that AB is a diameter and $\angle ADC={130}^{\circ }$ . Find $\angle BAC$.

Answer 1

Draw a quadrilateral ABCD inscribed in a circle having centre O.
Given, $\angle ADC={130}^{\circ }$
Since ABCD is a quadrilateral inscribed in a circle, ABCD becomes a cyclic quadrilateral.

$\therefore$ The sum of opposite angles of a cyclic quadrilateral is ${180}^{\circ }$.
$\Rightarrow \angle ADC+\angle ABC={180}^{\circ }$
$\Rightarrow {130}^{\circ }+\angle ABC={180}^{\circ }$
$\Rightarrow \angle ABC={50}^{\circ }$
Since, AB is a diameter of a circle, the angle subtended by AB to the circle is right angle.
$\therefore \angle ACB={90}^{\circ }$

$In\Delta ABC,\angle BAC+\angle ACB+\angle ABC={180}^{\circ }$ [by angle sum property of a triangle]
$\angle BAC+{90}^{\circ }+{50}^{\circ }={180}^{\circ }$
$\Rightarrow \angle BAC={180}^{\circ }-({90}^{\circ }+{50}^{\circ })$
$\Rightarrow \angle BAC={180}^{\circ }-{140}^{\circ }={40}^{\circ }$