Question

In the given figure, if $\angle ACB:\angle ABC=1:4,$
then ∠OAD is

• A.
  $5^{\circ }$
• B.
  ${75}^{\circ }$
• C.
  ${50}^{\circ }$
• D.
  ${30}^{\circ }$

Answer 1

The correct option is D

${30}^{\circ }$
Given:$\frac{\angle ACB}{\angle ABC}=\frac{1}{4}.........\left(i\right)$


Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

$\therefore \angle AOB=2\angle ACB\Rightarrow \angle ACB=\frac{{60}^{\circ }}{2}={30}^{\circ }\dots ..\left(ii\right)\text{From (i), we get}\angle ABC=4\angle ACB=4\times {30}^{\circ }\text{[From (ii)]}={120}^{\circ }\dots ..\left(iii\right)Now,OA=OB=Radii\Rightarrow \angle OAB=\angle OBAIn\Delta OAB,\text{by Angle sum property,}\therefore \angle OAB+\angle OBA+{60}^{\circ }={180}^{\circ }\Rightarrow 2\angle OBA={120}^{\circ }(\because \angle OAB=\angle OBA)\Rightarrow \angle OBA={60}^{\circ }\dots ..\left(iv\right)Now,\angle OBC=\angle ABC–\angle OBA={120}^{\circ }–{60}^{\circ }={60}^{\circ }\text{[From (iii) and (iv)]}In\Delta ABC,\text{by Angle sum property,}\angle CAB+\angle ABC+\angle ACB={180}^{\circ }\Rightarrow \angle CAB+{120}^{\circ }+{30}^{\circ }={180}^{\circ }\text{[From (ii) and (iii)]}\Rightarrow \angle CAB={180}^{\circ }–{150}^{\circ }={30}^{\circ }\dots ..\left(v\right)\therefore \angle OAD=\angle OAB–\angle CAB={60}^{\circ }–{30}^{\circ }={30}^{\circ }\text{[From (iv) and (v)]}$

Hence, the correct answer is option (d).