In the figure, DFH and EFG are straight lines. FD = FE and GF = GH. If ∠EDF = 50°, find
- ∠DFG
- ∠FGH
(a)
∠DFG
= 50° + 50°
= 100° (Exterior angle of a triangle)
(b)
∠GFH
= 180° - 100°
= 80° (Angles on a straight line)
∠FGH
= 180° - 80° - 80°
= 20° (Isosceles triangle)
Answer(s): (a) 100°; (b) 20°