Question

In the given circle, with AB as a diameter, find the value of x.

Answer 1

$\angle ADB={90}^{\circ }\text{[Angle in a semi-circle]}$
$\angle ABD=\angle ACD={30}^{\circ }\text{[Angles in the same segment]}$
In $△ADB$, we have
$\angle BAD+\angle ADB+\angle ABD={180}^{\circ }\text{(Sum of angles of a triangle is}{180}^{\circ })$
$x+{90}^{\circ }+{30}^{\circ }={180}^{\circ }$
$\Rightarrow x=({180}^{\circ }-{120}^{\circ })={60}^{\circ }$
$\text{Hence},x={60}^{\circ }$