In the figure, EFGH and FKLM are two rhombuses. If ∠KLM = 121° and ∠EHG = 102°, calculate
- ∠LKM
- ∠LGF
- ∠MNG
(a)
∠LKM
= (180° - 121°) ÷ 2
= 29.5° (Isosceles triangle)
(b)
∠LGF = ∠EHG = 102° (Corresponding angles)
(c)
∠MNG
= 102° - 29.5°
= 72.5° (Exterior angle of a triangle)
Answer(s): (a) 29.5°; (b) 102°; (c) 72.5°