In Figure (1), O is the centre of the circle. If ∠OAB = 40° and ∠OCB = 30°, find ∠AOC.

Question:

(i) In Figure (1), O is the centre of the circle. If OAB = 40° and ∠OCB = 30°, find ∠AOC.
(ii) In Figure (2), AB and C are three points on the circle with centre O such that ∠AOB = 90° and ∠AOC = 110°. Find ∠BAC.

 

Solution:

(i)  Join BO.

In ΔBOC, we have:
OC = OB (Radii of a circle)
⇒ OBC = OCB
OBC = 30°                 ...(i)
In ΔBOA, we have:
OB = OA   (Radii of a circle)
OBA = OAB    [∵ OAB = 40°]
OBA = 40°           ...(ii)
Now, we have:

ABC = OBC + OBA
= 30° + 40°    [From (i) and (ii)]
∴ ABC = 70°
The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.
i.e., AOC = 2ABC
= (2 × 70°) = 140°

(ii)

Here, $\angle B O C=\left\{360^{\circ}-\left(90^{\circ}+110^{\circ}\right)\right\}$

$=\left(360^{\circ}-200^{\circ}\right)=160^{\circ}$

We know that BOC = 2BAC

$\Rightarrow \angle B A C=\frac{\angle B O C}{2}=\left(\frac{160^{\circ}}{2}\right)=80^{\circ}$

Hence, BAC = 80°

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