In the diagram, PR is a diameter of the circle centre O. RS is a tangent at R and QPR = 58o. Find
-
A.
112o -
B.
116o -
C.
122o -
D.
148o
Correct Answer: Option C
Explanation
PRQ = 90 – 58 = 32o(angle in a semi-circle)
Since PRS = 90o(radius angular to tangent)
QRS = 90 + 32
= 122o