In the given figure, PQ is the diameter of the circle whose center is O. Given, ∠ROS = 42°, calculate ∠RTS.

Join PS.

∠PSQ = 90° [Angle in semi-circle is a right angle.]

We know that,

Angle at the centre is double the angle at the circumference subtended by the same chord.

⇒ ∠ROS = 2∠SPR

⇒ ∠SPR = $\dfrac{1}{2}$∠ROS = $\dfrac{42}{2}$ = 21°

From figure,

⇒ ∠SPT = ∠SPR = 21°.

From figure,

⇒ ∠PSQ = 90° [Angle in a semi-circle is a right angle.]

Since, QST is a straight line.

⇒ ∠PSQ + ∠PST = 180°

⇒ 90° + ∠PST = 180°

⇒ ∠PST = 90°.

In △PST,

⇒ ∠PTS + ∠PST + ∠SPT = 180° [Angle sum property of triangle]

⇒ ∠PTS + 90° + 21° = 180°

⇒ ∠PTS + 111° = 180°

⇒ ∠PTS = 180° - 111° = 69°.

From figure,

∠RTS = ∠PTS = 69°.

Hence, ∠RTS = 69°.