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 = 50° and ∠JFK = 33°. Find
- ∠CFE
- ∠FHG
(a)
∠CFK
= (180° - 50°) ÷ 2
= 65° (Isosceles triangle)
∠CFE
= 180° - 65°
= 115° (Angles on a straight line)
(b)
∠HFG
= 65° - 33°
= 32°
∠FHG
= (180° - 32°) ÷ 2
= 74 ° (Isosceles triangle)
Answer(s): (a) 115°; (b) 74°