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 = 52° and ∠JFK = 31°. Find
- ∠CFE
- ∠FHG
(a)
∠CFK
= (180° - 52°) ÷ 2
= 64° (Isosceles triangle)
∠CFE
= 180° - 64°
= 116° (Angles on a straight line)
(b)
∠HFG
= 64° - 31°
= 33°
∠FHG
= (180° - 33°) ÷ 2
= 73.5 ° (Isosceles triangle)
Answer(s): (a) 116°; (b) 73.5°