In the figure, EFGH and FKLM are two rhombuses. If ∠KLM = 119° and ∠EHG = 101°, calculate
- ∠LKM
- ∠LGF
- ∠MNG
(a)
∠LKM
= (180° - 119°) ÷ 2
= 30.5° (Isosceles triangle)
(b)
∠LGF = ∠EHG = 101° (Corresponding angles)
(c)
∠MNG
= 101° - 30.5°
= 70.5° (Exterior angle of a triangle)
Answer(s): (a) 30.5°; (b) 101°; (c) 70.5°