In the figure, EFGH and FKLM are two rhombuses. If ∠KLM = 113° and ∠EHG = 95°, calculate
- ∠LKM
- ∠LGF
- ∠MNG
(a)
∠LKM
= (180° - 113°) ÷ 2
= 33.5° (Isosceles triangle)
(b)
∠LGF = ∠EHG = 95° (Corresponding angles)
(c)
∠MNG
= 95° - 33.5°
= 61.5° (Exterior angle of a triangle)
Answer(s): (a) 33.5°; (b) 95°; (c) 61.5°