The figure is not drawn to scale. YBDH is a parallelogram. AEG and ZCF are triangles. ∠AEF = 82° and FZC = 50°. Find
- ∠ZAE
- the sum of ∠ZFC, ∠AWC and ∠CXE.
(a)
∠ZAE
= 180° - 82°
= 98° (Interior angles, YB//HD)
(b)
∠AWC = ∠ZWX (Vertically opposite angles)
∠CXE = ∠FXW (Vertically opposite angles)
ZFXW is a quadrilateral.
∠ZFC + ∠FXW + ∠ZWX + 50° = 360° (Sum of angles in a quadrilateral)
∠ZFC + ∠FXW + ∠ZWX
= 360° - 50°
= 310°
Answer(s): (a) 98°; (b) 310°