Question

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

Answer 1

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

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

Thus, ∠OBQ=90

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Now, ∠ABQ+∠OBA=90

∠ABQ+35=90

∠ABQ=55

∘

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Answer 2

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