In the figure, 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: OT = OS = r and OP = 2r
In ΔTOP, sin∠TPO = TO/OP = r/2r = 1/2
Since, sin 30° = 1/2
Therefore, ∠TPO = 30°
Similarly for ∠OPS = 30°
Now, ∠TPS = ∠TPO + ∠OPS = 30° + 30° = 60°
As we know that ∠TPS + ∠TOS = 180°
So, ∠TOS = 180° - ∠TPS = 180° - 60° = 120°
Now, in ΔTOS, let ∠OST = ∠OTS = x°
Also, ∠TOS + x° + x° = 180°
120° + 2x° = 180°
2x° = 60°
x° = 30°
Therefore, ∠OST = ∠OTS = 30°.