Question

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

Answer 1

Considering the theorm,

•The angle subtended by an arc at the

centre is double the angle

subtended by it at any point on the

remaining part of the circle.

=><ROS=2<RQS ( By the above theorm)

=> <RQS =20°

Since , PQ is a diameter

=> <POQ =180°

also <POQ=2<PRQ (By the above

theorm)

=> <PRQ = 90°.

Now, <RQS+<PRQ+<RTS = 180°( Angle

sum property of triangle)

=> 20°+90°+<RTS=180°

=> <RTS+110°=180°

=> < RTS =70°

Answer 2

The value of ∠RTS is 70°

Explanation:

Given that O is the centre of a circle

Also, given that PQ is a diameter and ∠ROS=40°

We need to determine the ∠RTS

The degree measure theorem states that "angle subtended by an arc of a circle at the centre is twice the angle subtended by it at any point on the circle".

Thus, by degree measure theorem, we have,

$\angle R Q S=\frac{1}{2} \angle R O S$

$\angle R Q S=\frac{1}{2}(40)$

$\angle R Q S=20^{\circ}$

Let us consider the right angled triangle TRQ

By angle sum property, we get,

$\angle R Q T+Q R T+\angle R T S=180^{\circ}$

Substituting the values of these angles, we have,

$20^{\circ}+90^{\circ}+\angle R T S=180^{\circ}$

       $110^{\circ}+\angle R T S=180^{\circ}$

                  $\angle R T S=180^{\circ}-110^{\circ}$

                  $\angle R T S=70^{\circ}$

Thus, the value of ∠RTS is 70°

Learn more:

(1) PQ is diameter,PRT and QST are secants. Given angle ROS = 42 find angle RTS

brainly.in/question/2648718

(2) 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.

brainly.in/question/291998