In a circle with center O, PQ and RS are two chords. PQ and RS, when produced, meet at point K. If PR is a diameter and ∠ PKS = 66°, what is the value of ∠ QOS?

Correct option is B

Given:
PR is a diameter of the circle.
PQ and RS are chords that meet at point K when produced.
Angle PKS = 66°
Formula Used:
Angle between two secants intersecting outside the circle = $$\frac{1}{2}$$​ × (Difference of intercepted arcs)
Solution: 

Since PR is the diameter, the measure of arc PR is 180°.
The intercepted arcs for secants KQP and KSR are arc PR and arc QS.
Angle PKS = $$\frac{arc PR - arc QS}{2}$$​​
Substitute the known values into the formula.
​$$66 = \frac{180 - arc QS}{2}$$​​
132 = 180 - arc QS
arc QS = 180 - 132
arc QS = 48
The central angle subtended by an arc is equal to the degree measure of that arc.
Angle QOS = 48°
Final Answer
So the correct answer is (b)