Correct option is: (B) 2∠OPQ
\(\because\) Tangents drawn from a fixed outer point to circle are of equal length.
\(\therefore\) TP = TQ
and OP = OQ = radii of circle
and OT = OT (Common side)
\(\therefore\) \(\triangle\)OTP \(\cong\) \(\triangle\)OTQ (By sss congruence criteria)
\(\therefore\) \(\angle\)TOP = \(\angle\)TOQ......(1)
and \(\angle\) OTP = \(\angle\)OTQ (By C.P.C.T)
= \(\angle\) OTP + \(\angle\) OTQ = \(\angle\)PTQ
= \(\angle\)PTQ = 2 \(\angle\)OTP .....(2) (\(\because\)\(\angle\) OTP = \(\angle\)OTQ)
\(\because\) \(\angle\)OPT = 90°....(3) (Angle between tangent and drawn radius at point of contact)
In triangles \(\triangle\)OPR & \(\triangle\) ORQ
OP = OQ (Radii of circle)
OR = OR (Common side)
\(\angle\)TOP = \(\angle\)TOQ (From (1))
\(\therefore\) \(\triangle\) OPR \(\cong\)\(\triangle\)ORQ (By SAS congruence criteria)
\(\therefore\) \(\angle\)ORP = \(\angle\)ORQ
But \(\angle\)ORP & \(\angle\)ORQ will form a linear pair.
\(\therefore\) \(\angle\)ORP = \(\angle\)ORQ = \(\frac {180^{\circ}}2 = 90°\) ....(4)
Now, in \(\triangle\)OPT & \(\triangle\) ORP
\(\angle\)OPT = \(\angle\)ORP = 90° (From (3) & (4))
\(\angle\)TOP = \(\angle\)ROP (Common angle)
\(\therefore\) \(\triangle\)OPT \(\sim\) \(\triangle\)ORP (By AA similarity rule)
\(\therefore\) \(\angle\)OTP = \(\angle\)OPR = \(\angle\)OPQ (By property of similar triangles)
From (2), we have
\(\angle\)PTQ = 2 \(\angle\)OTP = 2 \(\angle\)OPQ
