**Q) **

In figure, PQ is a tangent from an external point P to a circle with centre O and OP cuts the circle at T and QOR is a

diameter. If ∠POR = 130° and S is a point on the circle, find ∠1 +∠2

**Ans: **

In the given diagram, it is given that: ∠ POR = 130^{0}

We know that the Angle subtended at the center by same arc is half than that of center,

∴ ∠TOR = 2 ∠2

∴ 130^{0 }= 2 ∠2

∴ ∠2 = 65^{0 }**…… equation (1)**

∵ ∠ROT = 130^{0 }

∴ ∠QOT = 180 – 130 = 50^{0 }

Also we can see that, ∠POQ = ∠QOT = 50^{0 }.**….. equation (2)**

We know that the angle between the radius and the tangent is 90^{0 }at the point of contact,

therefore ∠ PQO = 90^{0}.**….. equation (3)**

In Δ POQ, ∠1 + ∠POQ + ∠PQO = 180^{0 }

By putting values from equations (2) & (3), we get:

∠1 + ∠POQ + ∠PQO = 180^{0 }

∴ ∠1 + 50^{0} + 90^{0} = 180^{0 }

∴ ∠1 = 40^{0 }.**…. equation (4)**

By adding equations (1) and (4), we get:

∠ 1 + ∠ 2 = 40^{0 } + 65^{0 }

∴ **∠ 1 + ∠ 2 = 105 ^{0}**