Question

In Fig. 3, 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°.

Answer 1

AP is the tangent to the circle

OA _|_ AP (Radius is perpendicular to the tangent at the point of contact)

∠OAP = 90 degrees

In angle OAP, 

Sin ∠OPA= OA/OP = R/2R = 1/2

∠OPA = 30

In Angle ABP,

AP = BP

∠PAB = ∠PBA

so 60+ ∠PAB + ∠PBA = 180

60+2 ∠PAB = 180

∠PAB = 180 - 60/2

∠PAB = 60

But

as ∠OAP = OBP = 90

OAP = OBP

so,

60 + x = 90

x = 30

therefore,

∠OTS = OST= 30