Question

In the given figure if ‘O’ is centre of the circle, then find the value of x?

(a) $ {{68}^{\circ }} $
(b) $ {{63}^{\circ }} $
(c) $ {{252}^{\circ }} $
(d) None of these

Answer 1

Hint: We start solving the problem by find the angles subtended at the circumference and at centre by the arc AB. We then recall the fact that the angle inscribed at the centre of the circle by an arc is twice the angle subtended at the circumference by the same arc for the obtained angles. We then divide the angle subtended at the centre with 2 to get the required value of x.

Complete step by step solution:
According to the problem, we need to find the value of x from the given figure if ‘O’ is the centre of the circle.
Let us redraw the given figure.

We can see that the angle $ \angle ACB $ is the angle subtended at the circumference by the arc AB and the angle $ \angle AOB $ is the angle subtended at the centre by the arc AB.
We know that the angle inscribed at the centre of the circle by an arc is twice the angle subtended at the circumference by the same arc.
So, we have $ \angle AOB=2\times \angle ACB $ .
 $ \Rightarrow {{126}^{\circ }}=2\times x $ .
 $ \Rightarrow x=\dfrac{{{126}^{\circ }}}{2} $ .
 $ \Rightarrow x={{63}^{\circ }} $ .
So, we have found the value of x as $ {{63}^{\circ }} $ .
 $ \, therefore, $ The correct option for the given problem is (b).

Note:
We can also solve this problem as shown below:

From the figure, let us consider the triangle $ \Delta OBC $ .
We can see that OC and OB are the radii of the circle so, $ OC=OB $ .
We know that the angles opposite to the equal sides in a triangle are equal.
So, we get $ \angle OCB=\angle CBO $ .
We know that the sum of all angles in a triangle is $ {{180}^{\circ }} $ .
So, we get $ \angle OCB+\angle CBO+\angle BOC={{180}^{\circ }} $ .
 $ \Rightarrow 2\angle OCB+\angle BOC={{180}^{\circ }} $ .
 $ \Rightarrow \angle BOC={{180}^{\circ }}-2\angle OCB $ .
From the figure, we can see that $ \angle BOC+\angle BOD={{180}^{\circ }} $ .
So, we get $ \angle BOD+{{180}^{\circ }}-2\angle OCB={{180}^{\circ }} $ .
 $ \Rightarrow \angle BOD=2\angle OCB $ ---(1).
Now, let us consider the triangle $ \Delta OAC $ .
We can see that OA and OC are the radii of the circle so, $ OA=OC $ .
We know that the angles opposite to the equal sides in a triangle are equal.
So, we get $ \angle OCA=\angle CAO $ .
We know that the sum of all angles in a triangle is $ {{180}^{\circ }} $ .
So, we get $ \angle OCA+\angle CAO+\angle AOC={{180}^{\circ }} $ .
 $ \Rightarrow 2\angle OCA+\angle AOC={{180}^{\circ }} $ .
 $ \Rightarrow \angle AOC={{180}^{\circ }}-2\angle OCA $ .
From the figure, we can see that $ \angle AOC+\angle AOD={{180}^{\circ }} $ .
So, we get $ \angle AOD+{{180}^{\circ }}-2\angle OCA={{180}^{\circ }} $ .
 $ \Rightarrow \angle AOD=2\angle OCA $ ---(2).
From figure, we can see $ \angle AOD+\angle BOD=\angle AOB $ .
From equations (1) and (2), we get $ \angle AOB=2\angle OCA+2\angle OCB $ .
 $ \Rightarrow \angle AOB=2\left( \angle OCA+\angle OCB \right) $ .
From figure, we can see $ \angle OCA+\angle OCB=\angle ACB $ .
 $ \Rightarrow \angle AOB=2\left( \angle ACB \right) $ .
 $ \Rightarrow {{126}^{\circ }}=2x $ .
 $ \Rightarrow {{63}^{\circ }}=x $ .