In the figure, O Is the centre of the circle and GN is parallel to HK. LP = LN, ∠OHG = 58° and ∠PNL = 48°. Find
- ∠KHR
- ∠HKL
(a)
OG = OH = Radius
∠GHO = ∠HGO = 58° (Isosceles triangle, OGB)
∠HOR
= 58° + 58°
= 116° (Exterior angle of a triangle)
OH = OR = Radius
∠ORH
= (180° - 116°) ÷ 2
= 64° ÷ 2
= 32°
∠KHR = 32° (Alternate angles, HK//GE)
(b)
LP = LN
∠LNP = ∠LPN = 48° (Isosceles triangle LNF)
∠NLP
= 180° - 48° - 48°
= 84°
∠RLK = ∠NLP = 84° (Vertically opposite angles)
∠HKL
= 180° - 84°
= 96° (Interior angles)
Answer(s): (a) 32°; (b) 96°
