Q9 of 100 Page 17

In Fig. 16.138, O is the centre of a circle and PQ is a diameter. If ∠ROS=40°, find ∠RTS.


Since,

PQ is diameter


Then,


∠PRQ = 90o (Angle in semi-circle)


Therefore,


∠PRQ + ∠TRQ = 180o (Linear pair)


90o + ∠TRQ = 180o


∠TRQ = 90o


By degree measure theorem,


∠ROS = 2 ∠RQS


∠RQS = 20o


In triangle RQT, we have


∠RQT + ∠QRT + ∠RTS = 180o (By angle sum property)


20o + 90o + ∠RTS = 180o


∠RTS = 70o


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