Answer:bsdaosdfhaodsifhaosdfaidfhaoif
have a nice time reading and understanding his:)
Step-by-step explanation:
considering APB as the triangle
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, LAB = LOP + LOB = 30° + 30° = 60°
In APB,
PA = PB [lengths &tangents drawn from an external point to circle areequal]
⇒ ∠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.
