From a point P, two tangent …

the diameter of the circle , show that ∆APB is equilateral. Report by Abinaya20 01.04.2017 Answers THE BRAINLIEST ANSWER! swathika23 Swathika23 Ace ∠OAP = 90°(PA and PB are the tangents to the circle.) In ΔOPA, sin ∠OPA = OA/OP = r/2r [OP is the diameter = 2*radius] sin ∠OPA = 1 /2 = sin 30 ⁰ ∠OPA = 30° Similarly, ∠OPB = 30°. ∠APB = ∠OPA + ∠OPB = 30° + 30° = 60° In ΔPAB, PA = PB (tangents from an external point to the circle) ⇒∠PAB = ∠PBA ............(1) (angles opp.to equal sides are equal) ∠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° all angles are equal in an equilateral triangle.( 60 degrees) ΔPAB is an equilateral triangle hope it helps!!!!