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

Question

In Fig. 16.138, O is the centre of a circle and PQ is a diameter. If &ang; ROS =40&deg;, find &ang; RTS .

Philoid · Accepted Answer

Since,

PQ is diameter

Then,

∠PRQ = 90^o (Angle in semi-circle)

Therefore,

∠PRQ + ∠TRQ = 180^o (Linear pair)

90^o + ∠TRQ = 180^o

∠TRQ = 90^o

By degree measure theorem,

∠ROS = 2 ∠RQS

∠RQS = 20^o

In triangle RQT, we have

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

20^o + 90^o + ∠RTS = 180^o

∠RTS = 70^o