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 remaining part of the circle i.e. ∠APB is (where 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 ΔPOB, OP=OB since both are radii of the circle.
So, ∠OPB = ∠OBP
As exterior angle is equal to sum of interior opposite angles,
we can write ∠COB = ∠OPB + ∠OBP
= 2 ∠OPB
So, ∠COB = 2 ∠CPB
Similarly, in ΔPOA, OP=OA since both are radii of the circle.
So, ∠OPA = ∠OAP
As the exterior angle is equal to the sum of interior opposite angles,
we can write ∠COA = ∠OPA + ∠OAP
= 2 ∠OPA
So ∠COA = 2 ∠CPA
Hence, ∠COA+∠COB = 2(∠CPA + ∠CPB)
Or, ∠ AOB = 2∠APB
Substituting we can get ∠APB = 55∘.