Question

In the given figure op is equal to the diameter of a circle with centre o and p a and pb are the tangent prove that it is an equilateral triangle

Answer 1

Answer:

Step-by-step explanation

From a point P , two tangents PA and PB are drawn to a circle with ​centre O. If OP is the diameter of the circle , show that ∆APB is equilateral.

∠OAP = 90°(PA and PB are the tangents to the circle.)

In ΔOPA,

sin ∠OPA = OA/OP = r/2r [OP is the diameter = 2*radius]

sin ∠OPA = 1 /2 = sin 30 ⁰

∠OPA = 30°

Similarly,

∠OPB = 30°.

∠APB = ∠OPA + ∠OPB = 30° + 30° = 60°

In ΔPAB,

PA = PB (tangents from an external point to the circle)

⇒∠PAB = ∠PBA ............(1) (angles opp.to equal sides are equal)

∠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°

all angles are equal in an equilateral triangle.( 60 degrees)

ΔPAB is an equilateral triangle