In the figure, CDEF is a trapezium and CFGK is a rhombus. CF is parallel to DE and FG = FH. KFE and FJH are straight lines. ∠FCK = 48° and ∠JFK = 35°. Find
- ∠CFE
- ∠FHG
(a)
∠CFK
= (180° - 48°) ÷ 2
= 66° (Isosceles triangle)
∠CFE
= 180° - 66°
= 114° (Angles on a straight line)
(b)
∠HFG
= 66° - 35°
= 31°
∠FHG
= (180° - 31°) ÷ 2
= 74.5 ° (Isosceles triangle)
Answer(s): (a) 114°; (b) 74.5°