In the figure, O Is the centre of the circle and DH is parallel to EF. GK = GH, ∠OED = 53° and ∠KHG = 54°. Find
- ∠FEL
- ∠EFG
(a)
OD = OE = Radius
∠DEO = ∠EDO = 53° (Isosceles triangle, ODB)
∠EOL
= 53° + 53°
= 106° (Exterior angle of a triangle)
OE = OL = Radius
∠OLE
= (180° - 106°) ÷ 2
= 74° ÷ 2
= 37°
∠FEL = 37° (Alternate angles, EF//DE)
(b)
GK = GH
∠GHK = ∠GKH = 54° (Isosceles triangle GHF)
∠HGK
= 180° - 54° - 54°
= 72°
∠LGF = ∠HGK = 72° (Vertically opposite angles)
∠EFG
= 180° - 72°
= 108° (Interior angles)
Answer(s): (a) 37°; (b) 108°