In the figure, EFGH and FKLM are two rhombuses. If ∠KLM = 116° and ∠EHG = 100°, calculate
- ∠LKM
- ∠LGF
- ∠MNG
(a)
∠LKM
= (180° - 116°) ÷ 2
= 32° (Isosceles triangle)
(b)
∠LGF = ∠EHG = 100° (Corresponding angles)
(c)
∠MNG
= 100° - 32°
= 68° (Exterior angle of a triangle)
Answer(s): (a) 32°; (b) 100°; (c) 68°