Question

In the given figure , O is the centre of the circle and PQ is the diameter of the circle. If angle ROS = 50 degree , angle P = 60 degree,find angle RTS , RSQ and angle POR.

Answer 1

Since O is the centre and OP =OR =radius of the circle 
and ∠OPR=$60^{0}$ ∴∠ORP=$60^{0}$
and  ∠OPR+∠ORP+∠POR=$180^{0}$
⇒$60^{0}$+$60^{0}$+∠POR=$180^{0}$
∴∠POR=$60^{0}$
NEXT
OR=OR=Radius of the circle ∴ It becomes an isoscles triangle  and ∠ROS=$50^{0}$  ∴∠ORS=∠OSR=$65^{0}$ (Since the sum of angles in a triangle  is $180^{0}$
Since Points P,O ,Q are collinear ∴∠SOQ=$180^{0}$-($60^{0}$+$50^{0}$)=$70^{0}$
and OS=OQ =Radius of the circle ∴∠OSQ=∠OQS=$55^{0}$
∴∠RSQ=$65^{0}$+$55^{0}$=$120^{0}$

Answer 2

Answer:

ऐसे ऐसे ही दिन अटैचमेंट