# If tangents PA an

Given:

APB = 80°

Property 1: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 2: Sum of all angles of a quadrilateral = 360°.

Property 3: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By property 1,

PAO = 90°

PBO = 90°

By property 2,

APB + PAO + PBO + AOB = 360°

AOB = 360° - APB + PAO + PBO

AOB = 360° - (80° + 90° + 90°)

AOB = 360° - 260°

AOB = 100°

Now, in ∆POA and ∆POB

OA = OB [ radius of circle]

PA = PB [By property 3 (tangent from P)]

OP = OP [ common]

By SSS congruency,

∆POA POB

Hence, by CPCTC

POA = POB

Now,

AOB = 100°

POA + POB = 100° [∵∠AOB = POA + POB]

POA + POA = 100° [∵∠POA = POB]

2POA = 100°

POA = 50°

Hence, POA = 50°

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