In the figure, EFGH and FKLM are two rhombuses. If ∠KLM = 122° and ∠EHG = 93°, calculate
- ∠LKM
- ∠LGF
- ∠MNG
(a)
∠LKM
= (180° - 122°) ÷ 2
= 29° (Isosceles triangle)
(b)
∠LGF = ∠EHG = 93° (Corresponding angles)
(c)
∠MNG
= 93° - 29°
= 64° (Exterior angle of a triangle)
Answer(s): (a) 29°; (b) 93°; (c) 64°