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 = 43° and ∠KGL = 32°. Find
- ∠DGF
- ∠GJH
(a)
∠DGL
= (180° - 43°) ÷ 2
= 68.5° (Isosceles triangle)
∠DGF
= 180° - 68.5°
= 111.5° (Angles on a straight line)
(b)
∠JGH
= 68.5° - 32°
= 36.5°
∠GJH
= (180° - 36.5°) ÷ 2
= 71.75 ° (Isosceles triangle)
Answer(s): (a) 111.5°; (b) 71.75°