In the figure, EFGH and FKLM are two rhombuses. If ∠KLM = 106° and ∠EHG = 97°, calculate
- ∠LKM
- ∠LGF
- ∠MNG
(a)
∠LKM
= (180° - 106°) ÷ 2
= 37° (Isosceles triangle)
(b)
∠LGF = ∠EHG = 97° (Corresponding angles)
(c)
∠MNG
= 97° - 37°
= 60° (Exterior angle of a triangle)
Answer(s): (a) 37°; (b) 97°; (c) 60°