WAEC - Mathematics (2004 - No. 27)

In the diagram, KS is a tangent to the circle centre O at R and ∠ROQ = 80º. Find ∠QRS.

90^o

80^o

50^o

40^o

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 = 40º
∠QRS = 40º