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 = \(36^\circ\). Find ∠PQ...

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 ${}^{\mathrm{<br>}}$ (sum of angles in a triangle)

Since |PS| = |SR|; ∠SRP = ∠SPR

⇒ ∠PSR + ∠SRP + ∠SRP = 180o ${}^{\mathrm{<br>}}$

∠PSR + 2∠SRP = 180o ${}^{\mathrm{<br>}}$

36o ${}^{\mathrm{<br>}}$ + 2∠SRP = 180o ${}^{}$


2∠SRP = 180o ${}^{\mathrm{<br>}}$ - 36o ${}^{\mathrm{<br>}}$

2∠SRP = 144o

∠SRP = 144o2=720 $\frac{{\mathrm{<br>}}^{}}{}<br>$

∠SRP = ∠PQR (angle formed by a tangent and chord is equal to the angle in the alternate segment)

∴ ∠PQR = 720