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°
⇒ ∠OBP = 90°
In Δ OAP
sin θ =perpendicular/hypotenuse
Similarly ∠OPB = 30°
Now ∠APB =∠OPA + ∠OPB
= 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.
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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