The figure, not drawn to scale, O is the centre of the circle and LR // MQ // NP. Find
- ∠KON
- ∠KQO
(a)
∠ROM
= 180° - 102°
= 78° (Interior angles, LR//MO)
∠ONQ = ∠OQN = 31° (Isosceles triangle)
∠MON
= 31° + 31°
= 62° (Exterior angle of a triangle)
∠KON
= 78° + 62°
= 140°
(b)
∠QON
= 180° - 31° -31°
= 118° (Isosceles triangle)
∠KOQ
= 360° - 140° - 118°
= 102° (Angles at a point)
KO = OQ = Radius
∠KQO
= (180° - 102°) ÷ 2
= 78° ÷ 2
= 39° (Isosceles triangle)
Answer(s): (a) 140°; (b) 39°
