In the figure, DEFG and EJKL are two rhombuses. If ∠JKL = 122° and ∠DGF = 98°, calculate
- ∠KJL
- ∠KFE
- ∠LMF
(a)
∠KJL
= (180° - 122°) ÷ 2
= 29° (Isosceles triangle)
(b)
∠KFE = ∠DGF = 98° (Corresponding angles)
(c)
∠LMF
= 98° - 29°
= 69° (Exterior angle of a triangle)
Answer(s): (a) 29°; (b) 98°; (c) 69°