Question

In the given figure, O is the centre of the circle and AB is a chord of the circle. If
AOB = 110°, find <APB​

Answer 1

Step-by-step explanation:

figure, O is the centre of the circle and AB is a chord of the circle. If

AOB = 110°,

Answer 2

Answer:

∘

Given, ∠AOB=110

∘

In △AOB

OA=OB (Radius of the circle)

Thus, ∠OAB=∠OBA (Isosceles triangle property)

Sum of angles of the triangle = 180

∠AOB+∠OAB+∠OBA=180

110+2∠OBA=180

∠OBA=35

∘

Since, PQ is a tangent touching the circle at B.

Thus, ∠OBQ=90

∘

Now, ∠ABQ+∠OBA=90

∠ABQ+35=90

∠ABQ=55