In the figure, O Is the centre of the circle and GN is parallel to HK. LP = LN, ∠OHG = 52° and ∠PNL = 54°. Find
- ∠KHR
- ∠HKL
(a)
OG = OH = Radius
∠GHO = ∠HGO = 52° (Isosceles triangle, OGB)
∠HOR
= 52° + 52°
= 104° (Exterior angle of a triangle)
OH = OR = Radius
∠ORH
= (180° - 104°) ÷ 2
= 76° ÷ 2
= 38°
∠KHR = 38° (Alternate angles, HK//GE)
(b)
LP = LN
∠LNP = ∠LPN = 54° (Isosceles triangle LNF)
∠NLP
= 180° - 54° - 54°
= 72°
∠RLK = ∠NLP = 72° (Vertically opposite angles)
∠HKL
= 180° - 72°
= 108° (Interior angles)
Answer(s): (a) 38°; (b) 108°