CFGK and CEHK are parallelograms and DJH is an isosceles triangle. Given that ∠DHE = 37°, ∠CKJ = 63° and ∠FHG = 52°, find
- ∠EHF,
- ∠FEH.
(a)
∠DHJ
= (180° - 27°) ÷ 2
= 76.5° (Isosceles triangle)
∠EHF
= 180° - 76.5° - 37° - 52°
= 14.5° (Angles on a straight line)
(b)
∠FEH
= 180° - 52° -14.5°
= 113.5° (Interior angles)
Answer(s): (a) 14.5°; (b) 113.5°