In the diagram, O is the center of the circle QRS and ∠SQR = 28°. Find ∠ORS.
A.
\(56^0\)
B.
\(28^0\)
C.
\(76^0\)
D.
\(62^0\)
Correct answer is D
∠SOR = 2 × 28° = 56° (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°
