In fig. 3, PQ is a tangent from an external point P to a circle with center O and OP cuts the circle at T and QOR is a diameter. If and S is a point on the circle, find

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°