In the figure, DEFG and EJKL are two rhombuses. If ∠JKL = 124° and ∠DGF = 96°, calculate
- ∠KJL
- ∠KFE
- ∠LMF
(a)
∠KJL
= (180° - 124°) ÷ 2
= 28° (Isosceles triangle)
(b)
∠KFE = ∠DGF = 96° (Corresponding angles)
(c)
∠LMF
= 96° - 28°
= 68° (Exterior angle of a triangle)
Answer(s): (a) 28°; (b) 96°; (c) 68°