The figure, not drawn to scale, O is the centre of the circle and LR // MQ // NP. Find
- ∠KON
- ∠KQO
(a)
∠ROM
= 180° - 94°
= 86° (Interior angles, LR//MO)
∠ONQ = ∠OQN = 32° (Isosceles triangle)
∠MON
= 32° + 32°
= 64° (Exterior angle of a triangle)
∠KON
= 86° + 64°
= 150°
(b)
∠QON
= 180° - 32° -32°
= 116° (Isosceles triangle)
∠KOQ
= 360° - 150° - 116°
= 94° (Angles at a point)
KO = OQ = Radius
∠KQO
= (180° - 94°) ÷ 2
= 86° ÷ 2
= 43° (Isosceles triangle)
Answer(s): (a) 150°; (b) 43°