In the figure, O Is the centre of the circle and DH is parallel to EF. GK = GH, ∠OED = 56° and ∠KHG = 57°. Find
- ∠FEL
- ∠EFG
(a)
OD = OE = Radius
∠DEO = ∠EDO = 56° (Isosceles triangle, ODB)
∠EOL
= 56° + 56°
= 112° (Exterior angle of a triangle)
OE = OL = Radius
∠OLE
= (180° - 112°) ÷ 2
= 68° ÷ 2
= 34°
∠FEL = 34° (Alternate angles, EF//DE)
(b)
GK = GH
∠GHK = ∠GKH = 57° (Isosceles triangle GHF)
∠HGK
= 180° - 57° - 57°
= 66°
∠LGF = ∠HGK = 66° (Vertically opposite angles)
∠EFG
= 180° - 66°
= 114° (Interior angles)
Answer(s): (a) 34°; (b) 114°