In the figure, DEFG and EJKL are two rhombuses. If ∠JKL = 123° and ∠DGF = 96°, calculate
- ∠KJL
- ∠KFE
- ∠LMF
(a)
∠KJL
= (180° - 123°) ÷ 2
= 28.5° (Isosceles triangle)
(b)
∠KFE = ∠DGF = 96° (Corresponding angles)
(c)
∠LMF
= 96° - 28.5°
= 67.5° (Exterior angle of a triangle)
Answer(s): (a) 28.5°; (b) 96°; (c) 67.5°