In the figure, JCEH and JCDG are parallelograms. Given that ∠ECF = 13°, ∠JHG = 64° and JGH is an isosceles triangle where JG = JH. Find
- ∠CFE
- ∠JCF
(a)
∠CEH
= 180° - 64°
= 116° (Interior angles)
∠CFE
= 180° - 116° - 13°
= 51° (Angles sum of triangle)
(b)
∠JCF
= ∠CFE
= 51° (Alternate angles)
Answer(s): (a) 51°; (b) 51°