ADEH and ACFH are parallelograms and BGF is an isosceles triangle. Given that ∠BFC = 42°, ∠AHG = 61° and ∠DFE = 49°, find
- ∠CFD,
- ∠DCF.
(a)
∠BFG
= (180° - 22°) ÷ 2
= 79° (Isosceles triangle)
∠CFD
= 180° - 79° - 42° - 49°
= 10° (Angles on a straight line)
(b)
∠DCF
= 180° - 49° -10°
= 121° (Interior angles)
Answer(s): (a) 10°; (b) 121°