In the figure, DEFG is a trapezium and DGHL is a rhombus. DG is parallel to EF and GH = GJ. LGF and GKJ are straight lines. ∠GDL = 44° and ∠KGL = 28°. Find
- ∠DGF
- ∠GJH
(a)
∠DGL
= (180° - 44°) ÷ 2
= 68° (Isosceles triangle)
∠DGF
= 180° - 68°
= 112° (Angles on a straight line)
(b)
∠JGH
= 68° - 28°
= 40°
∠GJH
= (180° - 40°) ÷ 2
= 70 ° (Isosceles triangle)
Answer(s): (a) 112°; (b) 70°