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 = 45° and ∠KGL = 34°. Find
- ∠DGF
- ∠GJH
(a)
∠DGL
= (180° - 45°) ÷ 2
= 67.5° (Isosceles triangle)
∠DGF
= 180° - 67.5°
= 112.5° (Angles on a straight line)
(b)
∠JGH
= 67.5° - 34°
= 33.5°
∠GJH
= (180° - 33.5°) ÷ 2
= 73.25 ° (Isosceles triangle)
Answer(s): (a) 112.5°; (b) 73.25°