ADEH and ACFH are parallelograms and BGF is an isosceles triangle. Given that ∠BFC = 42°, ∠AHG = 65° and ∠DFE = 46°, find
- ∠CFD,
- ∠DCF.
(a)
∠BFG
= (180° - 23°) ÷ 2
= 78.5° (Isosceles triangle)
∠CFD
= 180° - 78.5° - 42° - 46°
= 13.5° (Angles on a straight line)
(b)
∠DCF
= 180° - 46° -13.5°
= 120.5° (Interior angles)
Answer(s): (a) 13.5°; (b) 120.5°