WAEC - Mathematics (1989 - No. 4)
In the diagram O is the center of the circle, if ∠QRS = 62o, find the value of ∠SQR.
14o
28o
31o
45o
90o
Selitys
< QSR = 90° (angle in a semi-circle)
\(\therefore\) < SQR = 180° - (90° + 62°)
= 28°
In the diagram O is the center of the circle, if ∠QRS = 62o, find the value of ∠SQR.
< QSR = 90° (angle in a semi-circle)
\(\therefore\) < SQR = 180° - (90° + 62°)
= 28°