It is given that PS and PT are tangents to the circle with centre O. Also, ∠SPT = 120°.
To prove: OP = 2PS
Proof:
In △PTO and △PSO,
PT = PS (Tangents drawn from an external point to a circle are equal in length.)
TO = SO (Radii of the circle)
∠PTO = ∠PSO = 90°
\(\therefore \triangle PTO \cong\triangle PSO\) (By SAS congruency)
Thus,
∠TPO = ∠SPO = \(\frac{120°}2 = 60° \)
Now,
In ∆PSO,
\(\cos 60° = \frac{PS}{OP}\)
⇒ \(\frac 12 = \frac{PS}{OP}\)
⇒ \(OP = 2PS\)
Hence proved.