Question

In the given figure, $O$ is the centre of the circle. The measure of $\angle ADB$ is

• A. ${90}^{\circ }$
• B. ${85}^{\circ }$
• C. ${95}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

The correct option is C ${95}^{\circ }$

$\angle CEB=\angle DEA={60}^{\circ }$ (vert. opp. $\angle s$)
In $△CEB,\angle ECB={180}^{\circ }-({60}^{\circ }+{25}^{\circ })$
$={180}^{\circ }-{85}^{\circ }={95}^{\circ }$ (Angle sum prop. of a $△$)
$\therefore$ $\angle ADB=\angle BCA\left(\angle BCE\right)={95}^{\circ }$ (Angles in the same segment)