The figure, not drawn to scale, O is the centre of the circle and LR // MQ // NP. Find
- ∠KON
- ∠KQO
(a)
∠ROM
= 180° - 100°
= 80° (Interior angles, LR//MO)
∠ONQ = ∠OQN = 29° (Isosceles triangle)
∠MON
= 29° + 29°
= 58° (Exterior angle of a triangle)
∠KON
= 80° + 58°
= 138°
(b)
∠QON
= 180° - 29° -29°
= 122° (Isosceles triangle)
∠KOQ
= 360° - 138° - 122°
= 100° (Angles at a point)
KO = OQ = Radius
∠KQO
= (180° - 100°) ÷ 2
= 80° ÷ 2
= 40° (Isosceles triangle)
Answer(s): (a) 138°; (b) 40°