In the figure, DEFG and EJKL are two rhombuses. If ∠JKL = 120° and ∠DGF = 101°, calculate
- ∠KJL
- ∠KFE
- ∠LMF
(a)
∠KJL
= (180° - 120°) ÷ 2
= 30° (Isosceles triangle)
(b)
∠KFE = ∠DGF = 101° (Corresponding angles)
(c)
∠LMF
= 101° - 30°
= 71° (Exterior angle of a triangle)
Answer(s): (a) 30°; (b) 101°; (c) 71°