EXAMPLE 31 From a point P, two tangentsPA and PB are drawn to a circle

1.	EXAMPLE 31 From a point P, two tangentsPA and PB are drawn to a circleC(O, r). lf OP-2, show that△APB is equilateral.[CBSE 2008, '11, '12]SOLUTIONLet OP meet the circle at Q
Answer» ∠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 Like my answer if you find it useful!

EXAMPLE 31 From a point P, two tangentsPA and PB are drawn to a circleC(O, r). lf OP-2, show that△APB is equilateral.[CBSE 2008, '11, '12]SOLUTIONLet OP meet the circle at Q

Answer»

∠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

Like my answer if you find it useful!