ADEH and ACFH are parallelograms and BGF is an isosceles triangle. Given that ∠BFC = 36°, ∠AHG = 65° and ∠DFE = 51°, find
- ∠CFD,
- ∠DCF.
(a)
∠BFG
= (180° - 22°) ÷ 2
= 79° (Isosceles triangle)
∠CFD
= 180° - 79° - 36° - 51°
= 14° (Angles on a straight line)
(b)
∠DCF
= 180° - 51° -14°
= 115° (Interior angles)
Answer(s): (a) 14°; (b) 115°