In the diagram above, O is the center of the circle, |SQ| = |QR| and ∠PQR = 68°. Calculate ∠PRS
-
34o
- 45o
- 56o
- 62o
- 68o
The correct answer is: A
Explanation
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)