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] ….[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]


Adding [4] and [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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