In the diagram above, O is the center of the circle, |SQ| = |QR| and ∠PQR = 68°. Calculate ∠PRS
A.
34o
B.
45o
C.
56o
D.
62o
E.
68o
Correct answer is A
From the figure, < PQR = 68°
\(\therefore\) < QRS = < QSR = \(\frac{180 - 68}{2}\) (base angles of an isos. triangle)
= 56°
\(\therefore\) < PRS = 90° - 56° = 34° (angles in a semi-circle)