O is the centre of the circle. AB is a minor arc of the circle. The angle subtended by AB at centre ∠AOB =110∘, then angle subtended by the arc at any point on the circle ∠APB is
[P is any point on the circle]
As P is any point on the circle, we can take that as shown in the figure.
Now, we can easily see that in △POB, OP = OB since both are radii of the circle.
So, ∠OPB =∠OBP
As external angle is equal to sum of interior opposite angles,
We can write ∠COB =∠OPB +∠OBP
So, ∠COB =2∠CPB
Similarly, in △POA, OP = OA since both are radii of the circle.
So, ∠OPA =∠OAP
As the external angle is equal to the sum of interior opposite angles.
We can write ∠COA =∠OPA +∠OAP
So, ∠COA =2∠CPA
Hence, ∠COA +∠COB =2(∠CPA +∠CPB)
Or, ∠ AOB =2∠APB
Substituting, we can get ∠APB =55∘.