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 = 50° and ∠KGL = 29°. Find
- ∠DGF
- ∠GJH
(a)
∠DGL
= (180° - 50°) ÷ 2
= 65° (Isosceles triangle)
∠DGF
= 180° - 65°
= 115° (Angles on a straight line)
(b)
∠JGH
= 65° - 29°
= 36°
∠GJH
= (180° - 36°) ÷ 2
= 72 ° (Isosceles triangle)
Answer(s): (a) 115°; (b) 72°