In the figure, O Is the centre of the circle and EK is parallel to FG. HL = HK, ∠OFE = 58° and ∠LKH = 48°. Find
- ∠GFN
- ∠FGH
(a)
OE = OF = Radius
∠EFO = ∠FEO = 58° (Isosceles triangle, OEB)
∠FON
= 58° + 58°
= 116° (Exterior angle of a triangle)
OF = ON = Radius
∠ONF
= (180° - 116°) ÷ 2
= 64° ÷ 2
= 32°
∠GFN = 32° (Alternate angles, FG//EE)
(b)
HL = HK
∠HKL = ∠HLK = 48° (Isosceles triangle HKF)
∠KHL
= 180° - 48° - 48°
= 84°
∠NHG = ∠KHL = 84° (Vertically opposite angles)
∠FGH
= 180° - 84°
= 96° (Interior angles)
Answer(s): (a) 32°; (b) 96°