In the figure, PQ is a chord of circle with center O and PT is a tange

1.	In the figure, PQ is a chord of circle with center O and PT is a tangent If ∠QPT = 60°, then find ∠PRQ.
Answer» From figure, OP ⊥ PT i.e. ∠OPT = 90° ∠OPQ = ∠OPT – ∠QPT ∠OPQ = 90° – 60° = 30° In ΔOPQ, (equal radii of circle) ∠OQP = ∠OPQ = 30° (Opposite angles are equal to equal sides) Now ∠OQP + ∠OPQ + ∠POQ = 180° Sum of angles = 30° + 30° + ∠POQ = 180° ∠POQ = 180° – 60° = 120° Exterior ∠POQ = 360° – 120° = 240° We know that angle subtended by centre of circle is twice the angle subtended on the circumference of circle. ∠POQ = 2∠PRQ ⇒ 240° = 2∠PRQ ⇒ ∠PRQ = \(\frac { { 240 }^{ \circ } }{ 2 } \) = 120° Hence ∠PRQ = 120°.

In the figure, PQ is a chord of circle with center O and PT is a tangent If ∠QPT = 60°, then find ∠PRQ.

Answer»

From figure,

OP ⊥ PT

i.e. ∠OPT = 90°

∠OPQ = ∠OPT – ∠QPT

∠OPQ = 90° – 60°

= 30°

In ΔOPQ, (equal radii of circle)

∠OQP = ∠OPQ = 30°

(Opposite angles are equal to equal sides)

Now

∠OQP + ∠OPQ + ∠POQ = 180°

Sum of angles = 30° + 30° + ∠POQ = 180°

∠POQ = 180° – 60° = 120°

Exterior ∠POQ = 360° – 120° = 240°

We know that angle subtended by centre of circle is twice the angle subtended on the circumference of circle.

∠POQ = 2∠PRQ

⇒ 240° = 2∠PRQ

⇒ ∠PRQ = \(\frac { { 240 }^{ \circ } }{ 2 } \) = 120°

Hence ∠PRQ = 120°.