The figure is not drawn to scale. SVXB is a parallelogram. UYA and TWZ are triangles. ∠UYZ = 80° and ZTW = 52°. Find
- ∠TUY
- the sum of ∠TZW, ∠UQW and ∠WRY.
(a)
∠TUY
= 180° - 80°
= 100° (Interior angles, SV//BX)
(b)
∠UQW = ∠TQR (Vertically opposite angles)
∠WRY = ∠ZRQ (Vertically opposite angles)
TZRQ is a quadrilateral.
∠TZW + ∠ZRQ + ∠TQR + 52° = 360° (Sum of angles in a quadrilateral)
∠TZW + ∠ZRQ + ∠TQR
= 360° - 52°
= 308°
Answer(s): (a) 100°; (b) 308°