In the figure, $O$ is the centre of a circle and $PQ$ is a diameter. If $\angle ROS = 40^o$, find $\angle RTS$.
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Given:

In the figure, $O$ is the centre of a circle and $PQ$ is a diameter.

$\angle ROS = 40^o$.

To do:

We have to find $\angle RTS$.

Solution:

Arc $RS$ subtends $\angle ROS$ at the centre and $\angle RQS$ at the remaining part of the circle.

Therefore,

$\angle RQS = \frac{1}{2} \angle ROS$

$= \frac{1}{2} \times 40^o$

$= 20^o$

$\angle PRQ = 90^o$              (Angle in a semi circle)

$\angle QRT = 180^o - 90^o = 90^o$             ($PRT$ is a straight line)

In $\triangle RQT$,

$\angle RQT + \angle QRT +\angle RTQ = 180^o$              (Angles of a triangle)

$20^o + 90^o + \angle RTQ = 180^o$

$\angle RTQ = 180^o - 110^o$

$\angle RTS = 70^o$

Hence $\angle RTS = 70^o$.

Tutorialspoint

Updated on: 10-Oct-2022

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