CFGK and CEHK are parallelograms and DJH is an isosceles triangle. Given that ∠DHE = 37°, ∠CKJ = 61° and ∠FHG = 49°, find
- ∠EHF,
- ∠FEH.
(a)
∠DHJ
= (180° - 22°) ÷ 2
= 79° (Isosceles triangle)
∠EHF
= 180° - 79° - 37° - 49°
= 15° (Angles on a straight line)
(b)
∠FEH
= 180° - 49° -15°
= 116° (Interior angles)
Answer(s): (a) 15°; (b) 116°