Question

In Fig. 10.85, if PR is tangent to the circle at P and Q is the centre of the circle, then ∠POQ =
A. 110°
B. 100°
C. 120°
D. 90°

Answer 1

Answer:

120

Step-by-step explanation:

PT is a tangent so ∠OPT=90

∠QPT =60

∠OPQ = ∠OPT-∠QPT

∠OPQ=90-60

∠OPQ=30

∠OPQ+∠OQP+POQ=180        (ANGLE SUM PROPERTY)

∠POQ=180-30-30

∠POQ=120

Angle OPQ= angle OQP = 30° ie.,

Angle POQ= 120°. Also, angle PRQ= ½ reflex angle POQ.

∠PRQ=1/2(360-120)

∠PRQ=240/2

∠PRQ=120

Hope it helps ..

Answer 2

Answer:

120° (option:- [c])

Step-by-step explanation:

Given,

PT is a tangent so ∠OPT=90  

∠QPT =60

w.k.t ∠OPQ = ∠OPT-∠QPT  

∠OPQ=90-60   =30

⇒∠OPQ+∠OQP+POQ=180                                                                                     (∵ Angle Sum Property:-sum of interior   angles of a triangle is 180°.)

⇒∠POQ=180-30-30

= 180 - 60

∴∠POQ=120

⇒∠OPQ= ∠OQP = 30°

⇒∠POQ= 120°. Also, ∠PRQ= ½ reflex ∠POQ.

∠PRQ=1/2 (360-120 =240/2

∴∠PRQ=120

Hope this solution helps you and a lot of other users