In the diagram, O is the center of the circle QRS and ∠SQR = 28°. Find ∠ORS.
Attached Figure / Diagram
A
\(56^0\)
B
\(28^0\)
C
\(76^0\)
D
\(62^0\)
Verified Correct Answer: D.
Step-by-Step Mathematical Solution
∠SOR = 2 \times 28^\circ = 56^\circ (angle at the centre is twice the angle at the circumference)
From ∆SOR
|OS| = |OR| (radii)
So, ∆SOR is isosceles.
∠ORS = \(\frac{180^0 - 56^0}{2} = \frac{124^0}{2}\) ( base angles of isosceles triangle are equal)
∴ ∠ORS = 62^\circ
Source: WAEC / JAMB / NECO Classroom Archive (Record #68888)