Question

Assertion (A) in the given figure, O is the centre of the circle with D, E and F as mid-points of AB, BO and OA respectively. If $\angle DEF$ is ${30}^{\circ }$, then $\angle ACB$ is ${60}^{\circ }$

Reason (R) Angle subtended by an arc at the centre is twice the angle subtended by it on the remaining part of the circle.

• A. (A) is true and (R) is the correct explanation of (A)
• B. (A) is false and (R) is the correct explanation of (A)
• C. A is true and (R) is false
• D. Both (A) and (R) are false

Answer 1

The correct option is A

(A) is true and (R) is the correct explanation of (A)

Given F, D and E are mid-point of sides OA, AB and OB, respectively.

By basic proportionality theorem,
FE||AB,
ED||OA
and FD||OB
We know that AFED is a parallelogram.
Now, $\angle DEF={30}^{\circ }\left[given\right]$
$\Rightarrow \angle FAD=\angle DEF={30}^{\circ }....\left(i\right)$ [opposite angles of a parallelogram are equal]
Now, OA=OB [radii of circle]
In $\Delta OAB,$
$\angle OAB=\angle OBA={30}^{\circ }$ [from Eq. (i)]
Now,
$\angle OAB+\angle OBA+\angle AOB={180}^{\circ }\Rightarrow {30}^{\circ }+{30}^{\circ }+\angle AOB={180}^{\circ }\therefore \angle AOB={120}^{\circ }$
Since angle subtended by an arc at the centre is twice the angle subtended by it on the remaining part of the circle,
$\therefore \angle ACB=\frac{1}{2}\angle AOB=\frac{1}{2}\times {120}^{\circ }$
Now, $\angle ACB={60}^{\circ }$
Reason (R) True