In the figure, $ P Q $ is a tangent from an external point $ P $ to a circle with centre $ O $ and $ O P $ cuts the circle at $ T $ and $ Q O R $ is a diameter. If $ \angle P O R=130^{\circ} $ and $ S $ is a point on the circle, find $ \angle 1+\angle 2 $."

Given:

In the figure, \( P Q \) is a tangent from an external point \( P \) to a circle with centre \( O \) and \( O P \) cuts the circle at \( T \) and \( Q O R \) is a diameter.

\( \angle P O R=130^{\circ} \) and \( S \) is a point on the circle.

To do: We have to find \( \angle 1+\angle 2 \).

 Solution:

Join $RT$.

$\angle POR = 130^o$

$\angle POQ = 180^o- \angle POR$ $= 180^o - 130^o$

$= 50^o$

$PQ$ is a tangent to the circle.

$\angle PQO = 90^o$

In $\triangle POQ$,

$\angle POQ + \angle PQO + \angle QPO = 180^o$

$50^o + 90^o + \angle 1 = 180^o$

$\angle 1 = 180^o - 140^o$

$\angle 1 = 40^o$

In $\triangle RST$,

$\angle RST = \frac{1}{2} \angle ROT$      (Angle subtended by an arc at the centre of a circle is twice the angle subtended at any point on the circumference)

$\angle 2 = \frac{1}{2} \times 130^o$

$= 65^o$

Therefore,

$\angle 1 + \angle 2 = 40^o + 65^o$

$= 105^o$

Therefore, \( \angle 1+\angle 2 = 105^o \).