WAEC - Mathematics (1992 - No. 23)
In the diagram above, O is the center of the circle QOR is a diameter and ∠PSR is 37°. Find ∠PRQ
37o
53o
65o
127o
147o
Објашњење
Construction: Join PQ.
Then < RSP = 37° = < RQP (angles on the same segment)
But < RPQ = 90° (angle in a semi-circle)
\(\therefore\) < PRQ = 180° - (90° + 37°)
= 53°