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 = 52° and ∠KGL = 29°. Find
- ∠DGF
- ∠GJH
(a)
∠DGL
= (180° - 52°) ÷ 2
= 64° (Isosceles triangle)
∠DGF
= 180° - 64°
= 116° (Angles on a straight line)
(b)
∠JGH
= 64° - 29°
= 35°
∠GJH
= (180° - 35°) ÷ 2
= 72.5 ° (Isosceles triangle)
Answer(s): (a) 116°; (b) 72.5°