The figure, not drawn to scale, O is the centre of the circle and JP // KN // LM. Find
- ∠HOL
- ∠HNO
(a)
∠POK
= 180° - 94°
= 86° (Interior angles, JP//KO)
∠OLN = ∠ONL = 33° (Isosceles triangle)
∠KOL
= 33° + 33°
= 66° (Exterior angle of a triangle)
∠HOL
= 86° + 66°
= 152°
(b)
∠NOL
= 180° - 33° -33°
= 114° (Isosceles triangle)
∠HON
= 360° - 152° - 114°
= 94° (Angles at a point)
HO = ON = Radius
∠HNO
= (180° - 94°) ÷ 2
= 86° ÷ 2
= 43° (Isosceles triangle)
Answer(s): (a) 152°; (b) 43°