In the figure, EFGH and FKLM are two rhombuses. If ∠KLM = 123° and ∠EHG = 94°, calculate
- ∠LKM
- ∠LGF
- ∠MNG
(a)
∠LKM
= (180° - 123°) ÷ 2
= 28.5° (Isosceles triangle)
(b)
∠LGF = ∠EHG = 94° (Corresponding angles)
(c)
∠MNG
= 94° - 28.5°
= 65.5° (Exterior angle of a triangle)
Answer(s): (a) 28.5°; (b) 94°; (c) 65.5°