From a point P, 2 tangents PA & PB are drawn to a circle with centre O.If OP=diameter of the circle show that triangle APB is equilateral.

gopal.mohanty... , Meritnation Expert added an answer, on 24/2/11

Here are the answers to your question.

The given information can be represented using a figure as

⇒ ∠OPA = 30°

Similarly, it can be proved that ∠OPB = 30°.

Now, ∠APB = ∠OPA + ∠OPB = 30° + 30° = 60°

In ∆PAB,

PA = PB [lengths of tangents drawn from an external point to a circle are equal]

⇒∠PAB = ∠PBA ...(1) [Equal sides have equal angles opposite to them]

∠PAB + ∠PBA + ∠APB = 180° [Angle sum property]

⇒∠PAB + ∠PAB = 180° – 60° = 120° [Using (1)]

⇒2∠PAB = 120°

⇒∠PAB = 60° ...(2)

From (1) and (2)

∠PAB = ∠PBA = ∠APB = 60°

∴ ∆PAB is an equilateral triangle.

Hope! This will help you.

Cheers!

*This conversation is already closed by Expert*

97% users found this answer helpful.