In the figure, EFGH and FKLM are two rhombuses. If ∠KLM = 117° and ∠EHG = 98°, calculate
- ∠LKM
- ∠LGF
- ∠MNG
(a)
∠LKM
= (180° - 117°) ÷ 2
= 31.5° (Isosceles triangle)
(b)
∠LGF = ∠EHG = 98° (Corresponding angles)
(c)
∠MNG
= 98° - 31.5°
= 66.5° (Exterior angle of a triangle)
Answer(s): (a) 31.5°; (b) 98°; (c) 66.5°