Question

the given figure, AP is tangent to circle with centre O at A. If angle BAP = 50°, then find angle AOB. ​

Answer 1

Answer:

Tangent is perpendicular to radius at the point of contact. That gives:

∠OPB=∠OPQ+∠QPB=∠OPQ+50

o

=90

o

⇒∠OPQ=40

o

.

Triangle OPQ is isosceles since OP=OQ. So ∠OPQ=∠OQP=40

o

.

Thus, ∠POQ=180−40−40=100

o

Answer 2

AP is tangent.

angle OAP = 90°

angle OAP = angle BPA + angle OAB

90° = 50 ° + OAB

angle OAB = 90° - 50°

angle OAB = 40°

Join BP.

angle OBP = angle OAP = 40°

OAB + ABO + AOB = 180°

180° - 40° + 40° = AOB

AOB = 100°

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