ADEH and ACFH are parallelograms and BGF is an isosceles triangle. Given that ∠BFC = 42°, ∠AHG = 64° and ∠DFE = 52°, find
- ∠CFD,
- ∠DCF.
(a)
∠BFG
= (180° - 27°) ÷ 2
= 76.5° (Isosceles triangle)
∠CFD
= 180° - 76.5° - 42° - 52°
= 9.5° (Angles on a straight line)
(b)
∠DCF
= 180° - 52° -9.5°
= 118.5° (Interior angles)
Answer(s): (a) 9.5°; (b) 118.5°