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 = 46° and ∠JFK = 30°. Find
- ∠CFE
- ∠FHG
(a)
∠CFK
= (180° - 46°) ÷ 2
= 67° (Isosceles triangle)
∠CFE
= 180° - 67°
= 113° (Angles on a straight line)
(b)
∠HFG
= 67° - 30°
= 37°
∠FHG
= (180° - 37°) ÷ 2
= 71.5 ° (Isosceles triangle)
Answer(s): (a) 113°; (b) 71.5°