In the figure, O Is the centre of the circle and GN is parallel to HK. LP = LN, ∠OHG = 59° and ∠PNL = 55°. Find
- ∠KHR
- ∠HKL
(a)
OG = OH = Radius
∠GHO = ∠HGO = 59° (Isosceles triangle, OGB)
∠HOR
= 59° + 59°
= 118° (Exterior angle of a triangle)
OH = OR = Radius
∠ORH
= (180° - 118°) ÷ 2
= 62° ÷ 2
= 31°
∠KHR = 31° (Alternate angles, HK//GE)
(b)
LP = LN
∠LNP = ∠LPN = 55° (Isosceles triangle LNF)
∠NLP
= 180° - 55° - 55°
= 70°
∠RLK = ∠NLP = 70° (Vertically opposite angles)
∠HKL
= 180° - 70°
= 110° (Interior angles)
Answer(s): (a) 31°; (b) 110°