Question

From a point P, two tangents PA and PB

are drawn to a circle with center 0. If OP is equal to diameter of the circle, show
that triangle APB is equilateral.​

Answer 1

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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.