In the figure, EFH is a triangle with FH = FE, while DEFG is a parallelogram and GFH is a straight line. Given ∠FGD = 96°, ∠CFH = 145° and ∠DCF = 54°, find
- ∠FEH
- ∠CFE
(a)
∠DGF
= ∠EFH
= 96° (Corresponding angles)
∠FEH
= (180° - 96°) ÷ 2
= 42° (Isosceles triangle)
(b)
∠CFE
= 360° - 145° - 96°
= 119° (Angles at a point)
Answer(s): (a) 42°; (b) 119°