ADEH and ACFH are parallelograms and BGF is an isosceles triangle. Given that ∠BFC = 36°, ∠AHG = 63° and ∠DFE = 48°, find
- ∠CFD,
- ∠DCF.
(a)
∠BFG
= (180° - 26°) ÷ 2
= 77° (Isosceles triangle)
∠CFD
= 180° - 77° - 36° - 48°
= 19° (Angles on a straight line)
(b)
∠DCF
= 180° - 48° -19°
= 113° (Interior angles)
Answer(s): (a) 19°; (b) 113°