Question

PA and PB are tangents to the circle with Centre o and PQ = OQ prove that triangle PAB is equilateral Triangle

Answer 1

PA and PB are the tangents to the circle.
∴ OA ⊥ PA
⇒ ∠OAP = 90°
In ΔOPA,
sin ∠OPA = OA OP = r 2r [Given OP is the diameter of the circle]
⇒ sin ∠OPA = 1 2 = sin 30 ⁰
⇒ ∠OPA = 30°
Similarly, it can be proved that ∠OPB = 30°.
Now, ∠APB = ∠OPA + ∠OPB = 30° + 30° = 60°
In ΔPAB,
PA = PB [length of tangents drawn from an external point to a circle are equal]
⇒∠PAB = ∠PBA ............(1) [Equal sides have equal angles opposite to them]
∠PAB + ∠PBA + ∠APB = 180° [Angle sum property]
⇒∠PAB + ∠PAB = 180° – 60° = 120° [Using (1)]
⇒2∠PAB = 120°
⇒∠PAB = 60° .............(2)
From (1) and (2)
∠PAB = ∠PBA = ∠APB = 60°
∴ ΔPAB is an equilateral triangle.