In the figure, not drawn to scale, FHJK and EGKL are rhombuses. Given that ∠JHG = 120° and ∠FEL = 124°, find
- ∠ELG
- ∠GWF.
(a)
∠ELG
= (180° - 124°) ÷ 2
= 56 ÷ 2
= 28° (Isosceles triangle, EL = ER)
(b)
∠GFK
= 180° - 120°
= 60° (Interior angles, JH//KQ)
∠LGE = ∠ELG = 28° (Isosceles Jriangle, EL = ER)
∠GWF
= 180° - 60° - 28°
= 92° (Angles sum of triangle, WGQ)
Answer(s): (a) 28°; (b) 92°