Question

please help me answer this question from lesson circles in math​

Answer 1

Step-by-step explanation:

Given:- PR and PS are tangent of a circle with centre O, where P is a point external to the circle.

To Prove:- ∠OPR = ∠ORQ

Proof:-

Now,

P is a point external to the circle,

Thus,

PR = PS

∴∠PRQ = ∠PSQ [Angles opposite to equal sides are equal] or [Properties of Isosceles Triangle] ---- 1

Now,

In ΔPRS,

∠PRQ + ∠PSQ + ∠RPS = 180° [Angle Sum Property]

From eq.1

∠PRQ + ∠PRQ + ∠RPS = 180°

2∠PRQ = 180° - ∠RPS

∠PRQ = (180° - ∠RPS)/2

∠PRQ = 90° - (∠RPS/2) ------ 1

Now, we know that,

OR ⊥ PR [Radius ⊥ Tangent at the point of contact]

Thus,

∠ORP = 90°

But,

∠ORP = ∠ORQ + ∠PRQ

90° = ∠ORQ + ∠PRQ

∴ ∠PRQ = 90° - ∠ORQ ----- 2

From eq.1 and eq.2 we get,

90° - ∠ORQ = 90° - (∠RPS)/2

- ∠ORQ = 90° - 90° - (∠RPS)/2

-∠ORQ = -∠RPS/2

Thus,

∠ORQ = ∠RPS/2

2∠ORQ = ∠RPS ------ 3

Now,

we know that,

A line drawn from the centre of a circle to an external point, bisects the angles formed by the two tangents,

Thus,

∠OPR = ∠OPS

Now,

∠RPS = ∠OPR + ∠OPS

∠RPS = ∠OPR + ∠OPR

∠RPS = 2∠OPR ------ 4

From eq.3 and eq.4 we get,

2∠ORQ = 2∠OPR

∠ORQ = (2∠OPR)/2

∴ ∠ORQ = ∠OPR

Hence proved

Hope it helped and believing you understood it........All the best