In the figure, DEFG and EJKL are two rhombuses. If ∠JKL = 114° and ∠DGF = 96°, calculate
- ∠KJL
- ∠KFE
- ∠LMF
(a)
∠KJL
= (180° - 114°) ÷ 2
= 33° (Isosceles triangle)
(b)
∠KFE = ∠DGF = 96° (Corresponding angles)
(c)
∠LMF
= 96° - 33°
= 63° (Exterior angle of a triangle)
Answer(s): (a) 33°; (b) 96°; (c) 63°