Answer :

Given : From an external point P, Two tangents PT and PS are drawn to a circle with center O and radius r and OP = 2r

To prove : ∠OTS = ∠OST = 30°

Proof :

OS = OT [radii of same circle]

∠OTS = ∠OST [angles opposite to equal sides are equal] ….[1]

As tangent drawn at a point on circle is perpendicular to the radius through point of contact.

OT⏊TP And OS⏊SP

∠OSP = 90°

∠OST + ∠PST = 90°

∠PST = 90° - ∠OST …[2]

In triangle PTS

PT = PS [ Tangents drawn from an external point to a circle are equal]

∠PST = ∠PTS = 90° - ∠OST [FROM 2]

IN △PTS

∠PTS + ∠PST + ∠SPT = 180° [angle sum property of a triangle]

90 - ∠OST + 90 - ∠OST + ∠SPT = 180

∠SPT = 2∠OST …[3]

In △OTP, OT⏊TP

∠OPT = 30° …[4]

Similarly, in △OSP

∠OPS = 30° …[5]

Adding [4] and [5]

∠OPT + ∠OPS = 30° + 30°

∠SPT = 60°

Now putting this value in [3]

∠SPT = 2∠OST

60 = 2∠OST

∠OST = 30° …[6]

From [1] and [6]

∠OST = ∠OTS = 30°

Hence proved.

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