Question

op is equal to the diameter of a circle with Centre O and P A and PB are the tangents 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°