# In Fig. 3,

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.

OTTP And OSSP

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 = 2OST …[3]

In OTP, OTTP

OPT = 30° …[4]

Similarly, in OSP

OPS = 30° …[5]

OPT + OPS = 30° + 30°

SPT = 60°

Now putting this value in [3]

SPT = 2OST

60 = 2OST

OST = 30° …[6]

From [1] and [6]

OST = OTS = 30°

Hence proved.

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