In the figure, DFH and EFG are straight lines. FD = FE and GF = GH. If ∠EDF = 52°, find
- ∠DFG
- ∠FGH
(a)
∠DFG
= 52° + 52°
= 104° (Exterior angle of a triangle)
(b)
∠GFH
= 180° - 104°
= 76° (Angles on a straight line)
∠FGH
= 180° - 76° - 76°
= 28° (Isosceles triangle)
Answer(s): (a) 104°; (b) 28°