In Fig, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If OP=2r, show that  ∠ OTS = ∠ OST = 30°.

Given
OP= 2r
∠OTP = 90°  (radius drawn at the point of contact is perpendicular to the tangent)
Now ,
In ΔOTP,
Sin∠ OPT = OT/OP = 1/2 = Sin 30^o
⇒ ∠ OPT = 30^o
therefore,
∠ TOP=60^o
∴ ΔOTP is a 30^o-60^o-90^o, right triangle.

In ΔOTS,
OT = OS   … (Radii of the same circle)
therefore,
ΔOTS is an isosceles triangle.

∴∠OTS = ∠OST … (Angles opposite to equal sides of an isosceles triangle are equal)

In ΔOTQ and ΔOSQ

OS = OT … (Radii of the same circle)
OQ = OQ ...(side common to both triangles)
∠OTQ = ∠OSQ … (angles opposite to equal sides of an isosceles triangle are
equal)

∴ ΔOTQ ≅ ΔOSQ … (By S.A.S)
∴ ∠TOQ = ∠SOQ = 60° … (C.A.C.T)
∴ ∠TOS = 120° … (∠TOS = ∠TOQ + ∠SOQ = 60° + 60° = 120°)
∴ ∠OTS + ∠OST = 180° – 120° = 60°
∴ ∠OTS = ∠OST = 60° ÷ 2 = 30°