Question

From a point P, two tangents PA and PB are drawn to a circle C (O,r). If OP=2r, show that triangle APB is equilateral.

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°
APB is an equilateral triangle

Answer 2

Given:

P is a point two tangents

AP and AB are drawn to a circle C (O. r)

To Given:

ΔAPB is an equilateral.

Solution:

OA ⊥ AP [radius is perpendicular at point P]

∠OAP = 90°

In ΔOAP

sin ∠OPA = OA/OP

                = r/2r [diameter]

sin ∠OPA = 1/2

                = sin 30°

∠OPA = 30°

Similarly, ∠OPA = 30°

∠APB = ∠OPB +∠OPP = 30° + 30° = 60°

In APPB,

PA = PB [tangents drawn from external point]

∠PAB = ∠PBA ..(i)  [equal sides have equal angles]

now, by angle sum property of a triangle

∠PAB + ∠PBA + ∠APB = 180°

                                     = 180° - 60° [using (i)]

                               2∠PAB = 120°

                                  ∠PAB = 60°

Therefore, APB is an equilateral triangle.