CFGK and CEHK are parallelograms and DJH is an isosceles triangle. Given that ∠DHE = 37°, ∠CKJ = 62° and ∠FHG = 49°, find
- ∠EHF,
- ∠FEH.
(a)
∠DHJ
= (180° - 25°) ÷ 2
= 77.5° (Isosceles triangle)
∠EHF
= 180° - 77.5° - 37° - 49°
= 16.5° (Angles on a straight line)
(b)
∠FEH
= 180° - 49° -16.5°
= 114.5° (Interior angles)
Answer(s): (a) 16.5°; (b) 114.5°