In the figure, O Is the centre of the circle and DH is parallel to EF. GK = GH, ∠OED = 60° and ∠KHG = 51°. Find
- ∠FEL
- ∠EFG
(a)
OD = OE = Radius
∠DEO = ∠EDO = 60° (Isosceles triangle, ODB)
∠EOL
= 60° + 60°
= 120° (Exterior angle of a triangle)
OE = OL = Radius
∠OLE
= (180° - 120°) ÷ 2
= 60° ÷ 2
= 30°
∠FEL = 30° (Alternate angles, EF//DE)
(b)
GK = GH
∠GHK = ∠GKH = 51° (Isosceles triangle GHF)
∠HGK
= 180° - 51° - 51°
= 78°
∠LGF = ∠HGK = 78° (Vertically opposite angles)
∠EFG
= 180° - 78°
= 102° (Interior angles)
Answer(s): (a) 30°; (b) 102°