Question

In the given figure, AB is the diameter of the circle with centre O. If $\angle COB={60}^{\circ },AC=AD,$ then $\angle ABD$ is equal to

• A.
  ${45}^{\circ }$
• B.
  ${30}^{\circ }$
• C.
  ${75}^{\circ }$
• D.
  ${60}^{\circ }$

Answer 1

The correct option is D

${60}^{\circ }$
AC = AD (Given)




$\angle COB={60}^{\circ }\left(Given\right)Let\angle ABDbex.\angle ADB={90}^{\circ }(\because \text{Angle in a semicircle is right angle})\Rightarrow \angle DAB={90}^{\circ }–x\text{(Angle sum property) …..(i)}$


Now, angle subtended by an arc at the centre is double the angle subtended by it at any point on the circumference.

$\therefore \angle COB=2\angle CAB\Rightarrow \angle CAB=\frac{\angle COB}{2}=\frac{{60}^{\circ }}{2}={30}^{\circ }........\left(ii\right)In\Delta ACBand\Delta ADB,AC=AD\left(Given\right)\angle ACB=\angle ADB={90}^{\circ }(\because \text{Angle in a semicircle is right angle})AB=AB\left(Common\right)\Rightarrow \Delta ACB\cong \Delta AOB\text{(By RHS congruence rule)}\therefore \angle CAB=\angle DAB\text{(By CPCT)}\Rightarrow {30}^{\circ }={90}^{\circ }–x\text{[From (i) and (ii)]}\Rightarrow x={90}^{\circ }–{30}^{\circ }={60}^{\circ }Thus,\angle ABDis{60}^{\circ }.$

Hence, the correct answer is option (d).