In the figure, JCEH and JCDG are parallelograms. Given that ∠ECF = 11°, ∠JHG = 75° and JGH is an isosceles triangle where JG = JH. Find
- ∠CFE
- ∠JCF
(a)
∠CEH
= 180° - 75°
= 105° (Interior angles)
∠CFE
= 180° - 105° - 11°
= 64° (Angles sum of triangle)
(b)
∠JCF
= ∠CFE
= 64° (Alternate angles)
Answer(s): (a) 64°; (b) 64°