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

(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 = 2∠ABC
= (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 = 2∠BAC

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

Hence, ∠BAC = 80°