The figure, not drawn to scale, O is the centre of the circle and JP // KN // LM. Find
- ∠HOL
- ∠HNO
(a)
∠POK
= 180° - 102°
= 78° (Interior angles, JP//KO)
∠OLN = ∠ONL = 30° (Isosceles triangle)
∠KOL
= 30° + 30°
= 60° (Exterior angle of a triangle)
∠HOL
= 78° + 60°
= 138°
(b)
∠NOL
= 180° - 30° -30°
= 120° (Isosceles triangle)
∠HON
= 360° - 138° - 120°
= 102° (Angles at a point)
HO = ON = Radius
∠HNO
= (180° - 102°) ÷ 2
= 78° ÷ 2
= 39° (Isosceles triangle)
Answer(s): (a) 138°; (b) 39°
