The figure is not drawn to scale. BEGL is a parallelogram. DHK and CFJ are triangles. ∠DHJ = 80° and JCF = 56°. Find
- ∠CDH
- the sum of ∠CJF, ∠DZF and ∠FAH.
(a)
∠CDH
= 180° - 80°
= 100° (Interior angles, BE//LG)
(b)
∠DZF = ∠CZA (Vertically opposite angles)
∠FAH = ∠JAZ (Vertically opposite angles)
CJAZ is a quadrilateral.
∠CJF + ∠JAZ + ∠CZA + 56° = 360° (Sum of angles in a quadrilateral)
∠CJF + ∠JAZ + ∠CZA
= 360° - 56°
= 304°
Answer(s): (a) 100°; (b) 304°