In the figure, O Is the centre of the circle and DH is parallel to EF. GK = GH, ∠OED = 58° and ∠KHG = 49°. Find
- ∠FEL
- ∠EFG
(a)
OD = OE = Radius
∠DEO = ∠EDO = 58° (Isosceles triangle, ODB)
∠EOL
= 58° + 58°
= 116° (Exterior angle of a triangle)
OE = OL = Radius
∠OLE
= (180° - 116°) ÷ 2
= 64° ÷ 2
= 32°
∠FEL = 32° (Alternate angles, EF//DE)
(b)
GK = GH
∠GHK = ∠GKH = 49° (Isosceles triangle GHF)
∠HGK
= 180° - 49° - 49°
= 82°
∠LGF = ∠HGK = 82° (Vertically opposite angles)
∠EFG
= 180° - 82°
= 98° (Interior angles)
Answer(s): (a) 32°; (b) 98°
