# From a point P, t AP is the tangent to the circle,

According to the theorem which states that tangent to a circle is perpendicular to the radius through the point of contact.

⇒ ∠OAP = 90°

Also ⇒ ∠OBP = 90°

In Δ OAP

sin θ =perpendicular/hypotenuse As  Similarly ∠OPB = 30°

Now ∠APB =∠OPA + ∠OPB
=  30°+30°
= 60° .... (1)
In Δ PAB,

As PA and PB are drawn from external point P,

By theorem which states that the lengths of the two tangents drawn from external point to a circle are equal.

⇒ PA=PB

Also ∠PAB = ∠PBA  .... (2)

As ∠PAB + ∠PBA + ∠APB = 180° (sum of angles of triangle)

∠PAB + ∠PBA = 180° - ∠APB

∠PAB + ∠PBA = 180° - 60°

⇒ 2∠PAB = 120°

⇒ ∠PAB = 60° .... (3)

From 1 and 2 and 3,

∠PAB = ∠PBA = ∠APB = 60°

Hence ΔPAB is an equilateral triangle.

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