Question

in a circle with center O the secant is
a.PQ. b.XY c.QR
d.AB​

Answer 1

Answer:

Answer

Correct option is

C

55

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Given, ∠AOB=110

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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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