Q. 183.9( 7 Votes )

# In Fig. 3,

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] ….

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 …

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 …

In OTP, OTTP   OPT = 30° …

Similarly, in OSP

OPS = 30° …

Adding  and 

OPT + OPS = 30° + 30°

SPT = 60°

Now putting this value in 

SPT = 2OST

60 = 2OST

OST = 30° …

From  and 

OST = OTS = 30°

Hence proved.

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