Complete Question is: If PT and PS are tangents and O is the center of the circle having OP = 2r

then show that ∠OST = OTS = 30^°

Solution:

Given, OT = OS = r

and OP = 2r

From the figure,

In ΔTOP

     sin ∠TPO = TO/OP

=> sin ∠TPO = r/2r

=> sin ∠TPO = 1/2

=> sin ∠TPO = sin 30

=> ∠TPO = 30

Similarly, ∠OPS = 30

Now, from the figute,

     ∠TPS = ∠TPO + ∠OPS

=> ∠TPS = 30 + 30

=> ∠TPS = 60

Again, ∠TOS + ∠TPS = 180

=> ∠TOS + 60 = 180

=> ∠TOS = 180 - 60

=> ∠TOS = 120

Now, in ΔTOS,

let ∠OST = OTS = x

Since, ∠OST + ∠OTS + ∠TOS = 180

=> x + x + 120 = 180

=> 2x + 120 = 180

=> 2x = 180 - 120

=> 2x = 60

=> x = 60/2

=> x = 30

So, ∠OST = OTS = 30^°