In the figure, O Is the centre of the circle and EK is parallel to FG. HL = HK, ∠OFE = 60° and ∠LKH = 50°. Find
- ∠GFN
- ∠FGH
(a)
OE = OF = Radius
∠EFO = ∠FEO = 60° (Isosceles triangle, OEB)
∠FON
= 60° + 60°
= 120° (Exterior angle of a triangle)
OF = ON = Radius
∠ONF
= (180° - 120°) ÷ 2
= 60° ÷ 2
= 30°
∠GFN = 30° (Alternate angles, FG//EE)
(b)
HL = HK
∠HKL = ∠HLK = 50° (Isosceles triangle HKF)
∠KHL
= 180° - 50° - 50°
= 80°
∠NHG = ∠KHL = 80° (Vertically opposite angles)
∠FGH
= 180° - 80°
= 100° (Interior angles)
Answer(s): (a) 30°; (b) 100°
