ADEH and ACFH are parallelograms and BGF is an isosceles triangle. Given that ∠BFC = 39°, ∠AHG = 62° and ∠DFE = 46°, find
- ∠CFD,
- ∠DCF.
(a)
∠BFG
= (180° - 21°) ÷ 2
= 79.5° (Isosceles triangle)
∠CFD
= 180° - 79.5° - 39° - 46°
= 15.5° (Angles on a straight line)
(b)
∠DCF
= 180° - 46° -15.5°
= 118.5° (Interior angles)
Answer(s): (a) 15.5°; (b) 118.5°