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