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 = 51° and ∠JFK = 29°. Find
- ∠CFE
- ∠FHG
(a)
∠CFK
= (180° - 51°) ÷ 2
= 64.5° (Isosceles triangle)
∠CFE
= 180° - 64.5°
= 115.5° (Angles on a straight line)
(b)
∠HFG
= 64.5° - 29°
= 35.5°
∠FHG
= (180° - 35.5°) ÷ 2
= 72.25 ° (Isosceles triangle)
Answer(s): (a) 115.5°; (b) 72.25°