Question

In the given figure, PQ is the diameter of the circle whose center is at O. If ∠ROS = 44° and OR is the bisector of ∠PRQ, then what is the value of ∠RTS?

(a) 46° 

(b) 64° 

(c) 69° 

(d) None of these

Answer 1

Answer : (d) None of these

Since OR isthe bisector of ∠PRQ 

∠PRO = ∠ORQ = 45° (∵ ∠PRQ = 90° ∠ in a semicircle)

Also, OP = OR (radii) 

∴  ∠OPR = ∠ORP = 45° 

In ∆ ORS, OR = OS ⇒ ∠ORS = ∠OSR = \(\frac{180° -44°}{2}\) = 68°

∴ ∠MRS = 68° – 45° = 23° 

⇒ ∠PRS = 90° + 23° = 113° 

∠PRS + ∠PQS = 180°

⇒ ∠PQS = 180° – ∠PRS 

= 180° – 113° = 67° (opp. ∠s of cyclic quad. PQSR) 

In ∆ PTQ, ∠PTQ = 180° – (∠QPT + ∠PQT) 

= 180° – (45° + 67°) = 68° 

⇒ ∠RTS = ∠PTQ = 68°.