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 = 34° (Isosceles triangle)
∠MON
= 34° + 34°
= 68° (Exterior angle of a triangle)
∠KON
= 78° + 68°
= 146°
(b)
∠QON
= 180° - 34° -34°
= 112° (Isosceles triangle)
∠KOQ
= 360° - 146° - 112°
= 102° (Angles at a point)
KO = OQ = Radius
∠KQO
= (180° - 102°) ÷ 2
= 78° ÷ 2
= 39° (Isosceles triangle)
Answer(s): (a) 146°; (b) 39°