In the diagram, KS is a tangent to the circle centre O at R and ∠ROQ = 80°. Find ∠QRS.
A.
90o
B.
80o
C.
50o
D.
40o
Correct answer is D
QOP = 180 - 100(straight line angle) → 80°
OPQ and OQP are isosceles of 40° each.
∠QRS = ∠RPQ (theorem: angle in the alternate segment are equal)
∠RPQ = 40°
∠QRS = 40°