In the figure, JCEH and JCDG are parallelograms. Given that ∠ECF = 11°, ∠JHG = 68° and JGH is an isosceles triangle where JG = JH. Find
- ∠CFE
- ∠JCF
(a)
∠CEH
= 180° - 68°
= 112° (Interior angles)
∠CFE
= 180° - 112° - 11°
= 57° (Angles sum of triangle)
(b)
∠JCF
= ∠CFE
= 57° (Alternate angles)
Answer(s): (a) 57°; (b) 57°