Question

in the adjoining figure PA and PB are tangents to a circle with center O if OP is equal to the diameter of the circle prove that ABP is an equilateral triangle

Answer 1

AP is the tangent to the circle.

∴ OA ⊥ AP  (Radius is perpendicular to the tangent at the point of contact)

⇒ ∠OAP = 90º 

In Δ OAP,

sin ∠OPA = OA/OP = r/2r [Diameter of the circle]

∴ sin ∠OPA = 1/2 = sin 30º

⇒ ∠OPA =  30º

Similarly, it can he prayed that ∠OPB = 30
How, LAPB = LOPP + LOPB = 30° + 30° = 60° 
In APPB, 
PA = PB [lengths &tangents drawn from an external point to circle are equal] 
⇒ ∠PAB = ∠PBA --- (1) [Equal sides have equal angles apposite to them] 
∠PAB + ∠PBA + ∠APB = 180° [Angle sum property] 

∠PAB + ∠PBA + ∠APB = 180° - 60° [Using (1)] 
⇒ 2∠PAB = 120°

⇒ ∠PAB = 60° 
From (1) and (2) 
∠PAB = ∠PBA = ∠APB = 60°
∆ABP is an equilateral triangle

Answer 2

Step-by-step explanation:We redraw the given figure by joining A to B as

shown below.

Since OA is radius and PA is tangent at A, OA AP =

. Now in right angle triangle Δ OAP, OP is equal to

diameter of circle, thus

OP = 2OA

OP

OA

2

1 =

sin θ 2

1 = & θ = 30c

Since PO bisect the angle +APB ,

Hence, +APB = = 2 30 60 # c º

Now, in TAPB ,

AP = AB

+PAB = +PBA

º 2

180  60 60 = − =

Thus TAPB is an equilateral triangle.