# In fig. 3, PQ is

Given:

POR = 130°

The inscribed angle theorem states that an angle θ inscribed in a circle is half of the central angle 2θ that subtends the same arc on the circle.

Property 1: Sum of all angles of a triangle = 180°

Property 2: Sum of all angles of a straight line = 180°

Property 3: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

By above property, PQO = 90°.

POR = 130°

TOR = 130°

Now,

TOR = 130°

By inscribed angle theorem, By property 2,

TOR + TOQ = 180°

TOR + TOQ = 180°

TOQ = 180° - TOR

TOQ = 180° - 130°

TOQ = 50°

Now By property 1,

PQO + QPO + POQ = 180°

QPO = 180° - (PQO + POQ)

QPO = 180° - (90° + 50°)

QPO = 180° - 140°

1 = QPO = 40°

Hence, 1 = 40° and 2=65°

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