Question

In the given figure, if AOB is a diameter and $\angle BCD={150}^{\circ },$ then the measure of $\angle AED$ is

• A.
  
  ${140}^{\circ }$
• B.
  ${70}^{\circ }$
• C.
  ${120}^{\circ }$
• D.
  ${75}^{\circ }$

Answer 1

The correct option is C

${120}^{\circ }$
Let $\angle BCOand\angle DCO$ be x and y respectively.

$\angle BCD=\angle BCO+\angle DCO\Rightarrow x+y={150}^{\circ }(\because \angle BCD={150}^{\circ })\dots ..\left(i\right)Now,\angle EAO=\angle AEO=z\left(say\right)\text{(OA = OE = Radii)}Also,\angle DEO=\angle ODE=w\left(say\right)\text{(OD = OE = Radii)}\therefore \angle DOE={180}^{\circ }–2w\text{(Angle sum property)}Now,\angle DOE=2\angle DAE\text{(Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle)}\therefore {180}^{\circ }–2w=2a[\angle DAE=a(say\left)\right]\Rightarrow 2(a+w)={180}^{\circ }\Rightarrow a+w=90°\dots ..\left(ii\right)Now,\angle OAD=\angle ODA=z–a(\because OD=OA=Radii)Also,\angle DAB+\angle DCB={180}^{\circ }(\because \text{ADCB is a cyclic quadrilateral})\therefore z–a+x+y={180}^{\circ }\Rightarrow z–a={180}^{\circ }–{150}^{\circ }={30}^{\circ }\dots ..\left(iii\right)\text{[From (i)]}\text{Adding (ii) and (iii), we get}a+w+z–a={90}^{\circ }+{30}^{\circ }\Rightarrow w+z={120}^{\circ }\therefore \angle AED={120}^{\circ }$

Hence, the correct answer is option (c).