P and Q are the middle points of two chords (not diameters) AB and AC respectively of a circle with centre at a point 0. The lines OP and OQ are produced to meet the circle respectively at the points R and S. T is any point on the major arc between the points R and S of the circle. If ∠BAC = 32°, ∠RTS = ?

In a circle radius passing through the mid-point of the chord is perpendicular to that chord.
Hence, OP and OQ are perpendicular to AB and AC respectively.
Hence, ∠APO and ∠AQO is 90° both.
In quadrilateral APOQ sum of all internal angles is 360°. i.e.
∠APO + ∠AQO + ∠PAQ + ∠POQ = 360°
Given, ∠BAC = 32°, thus 90° + 90° + 32° + ∠POQ = 360° ∴ ∠POQ = 148°
Also, ∠ROS = ∠POQ (∵The lines OP and OQ is extended to R and S) Or, ∠ROS = 148°
Now, since ∠ROS and ∠RTS have same arc RS but one subtend angle at centre of the circle while other at a point T anywhere on the remaining part of circle. Thus according to the
Theorem: The angle subtended by an arc at the centre of a circle is twice the angle subtended by the same arc at any other point on the remaining part of the circle.
∴ ∠ROS = 2∠RTS
∠RTS = 148/2
= 74°