In the figure, JCEH and JCDG are parallelograms. Given that ∠ECF = 14°, ∠JHG = 70° and JGH is an isosceles triangle where JG = JH. Find
- ∠CFE
- ∠JCF
(a)
∠CEH
= 180° - 70°
= 110° (Interior angles)
∠CFE
= 180° - 110° - 14°
= 56° (Angles sum of triangle)
(b)
∠JCF
= ∠CFE
= 56° (Alternate angles)
Answer(s): (a) 56°; (b) 56°