In the given figure, PQ is the diameter of a circle with center O. Two points R and S are chosen on the circle such that ∠ROS = 80◦. When PR and QS are extended, they meet at T. The value of ∠RTS is

The correct option is 50°.

Based on the provided image, we have a circle with center O and diameter PQ. Two points R and S lie on the circle, and the angle subtended by the arc RS at the center O is:

$\angle ROS=80°$

The line segments PR and QS are extended to meet at an external point T.

First, let us find the angle subtended by the arc RS at the point P on the circumference of the circle, which is:

$\angle RPS$

According to the circle theorem, the angle subtended by an arc at the center is twice the angle subtended by the same arc at any point on the remaining part of the circle. Therefore, we have:

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

Substituting the given value of 80° into the equation:

$\angle RPS=\frac{1}{2}\times 80°=40°$

Since the points P, R, and T lie on the same straight line, we also have:

$\angle SPT=\angle RPS=40°$

Next, since PQ is the diameter of the circle, the angle subtended by it in the semicircle is a right angle:

$\angle PSQ=90°$

Since the line segment QS is extended to T, the points Q, S, and T lie on a straight line. Thus, the angles ∠PSQ and ∠PST are supplementary:

$\angle PSQ+\angle PST=180°$

Substituting ∠PSQ = 90°:

$90°+\angle PST=180°$

$\angle PST=90°$

Now, consider the right-angled triangle PST. The sum of the interior angles in any triangle is 180°:

$\angle SPT+\angle PST+\angle PTS=180°$

Substituting ∠SPT = 40° and ∠PST = 90°:

$40°+90°+\angle PTS=180°$

$130°+\angle PTS=180°$

$\angle PTS=180°-130°=50°$

Since the line segments PR and QS meet at T, the angle ∠RTS is the same as ∠PTS. Therefore:

$\angle RTS=50°$