The figure, not drawn to scale, O is the centre of the circle and JP // KN // LM. Find
- ∠HOL
- ∠HNO
(a)
∠POK
= 180° - 96°
= 84° (Interior angles, JP//KO)
∠OLN = ∠ONL = 28° (Isosceles triangle)
∠KOL
= 28° + 28°
= 56° (Exterior angle of a triangle)
∠HOL
= 84° + 56°
= 140°
(b)
∠NOL
= 180° - 28° -28°
= 124° (Isosceles triangle)
∠HON
= 360° - 140° - 124°
= 96° (Angles at a point)
HO = ON = Radius
∠HNO
= (180° - 96°) ÷ 2
= 84° ÷ 2
= 42° (Isosceles triangle)
Answer(s): (a) 140°; (b) 42°