The figure is not drawn to scale. YBDH is a parallelogram. AEG and ZCF are triangles. ∠AEF = 80° and FZC = 56°. Find
- ∠ZAE
- the sum of ∠ZFC, ∠AWC and ∠CXE.
(a)
∠ZAE
= 180° - 80°
= 100° (Interior angles, YB//HD)
(b)
∠AWC = ∠ZWX (Vertically opposite angles)
∠CXE = ∠FXW (Vertically opposite angles)
ZFXW is a quadrilateral.
∠ZFC + ∠FXW + ∠ZWX + 56° = 360° (Sum of angles in a quadrilateral)
∠ZFC + ∠FXW + ∠ZWX
= 360° - 56°
= 304°
Answer(s): (a) 100°; (b) 304°