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°
Answers
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:
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
