Question

A circle with centre.
O Point A is in the
exterior of the circle.
Line AP and line
AQ are tangents at
point P and point Q
respectively P-A-S, Q-A-R anglePAR = 130°. Find
angleAOP.

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Answer 1

Answer:

Join PQ.

PAB + PAQ = 180o (Linear pair)

PAQ + 125o = 180o

PAQ = 55o

Since the length of tangent from an external point to a circle are equal, PA = QA

Thus, in PAQ, APQ = AQP

In APQ,

APQ + AQP + PAQ = 180o (Angle sum property)

APQ + AQP + 55o = 180o

2APQ = 180o - 55o

APQ =

APQ = AQP =

Now, OQ and OP are radii of the circle, QA and PA are tangents, therefore, OQA = 90o and OPA = 90o

Now, OPQ + QPA = OPA = 90o

OPQ + = 90o

OPQ = 90o - ==

Similarly, OQP + PQA = OQA

OQP =

In POQ,

OQP + OPQ + POQ = 180o (Angle sum property)

+ + POQ = 180o

POQ = 180o - 55o = 125o