The diagram above is a circle with centre C. P, Q and S are points on the circumference. PS and SR are tangents to the circle. ∠PSR = 36o. Find ∠PQR
From ∆PSR |PS| = |SR| (If two tangents are drawn from an external point of the circle, then they are of equal lengths) ∴ ∆PSR is isosceles ∠PSR + ∠SRP + ∠SPR = 180o (sum of angles in a triangle) Since |PS| = |SR|; ∠SRP = ∠SPR ⇒ ∠PSR + ∠SRP + ∠SRP = 180o ∠PSR + 2∠SRP = 180o 36o + 2∠SRP = 180o 2∠SRP = 180o - 36o 2∠SRP = 144o
∠SRP = 144o2=720
∠SRP = ∠PQR (angle formed by a tangent and chord is equal to the angle in the alternate segment) ∴ ∠PQR = 720