The figure is not drawn to scale. YZBC is a parallelogram. ∠CYO is 59° and ∠BZA is 28°. YZA is an isosceles triangle and YZ = ZA.
- Find ∠OAZ.
- Find ∠YCB.
(a)
∠ZOY = 59° (Alternate angles, YC//ZB)
∠ZOA
= 180° - 59°
= 121° (Angles on a straight line)
∠OAZ
= 180° - 121° - 28°
= 31° (Angles sum of triangle)
(b)
∠ZYO = 31° (Isosceles triangle YZA)
∠YCB
= 180° - 59° - 31°
= 90° (Interior angles, YZ//CB)
Answer(s): (a) 31°; (b) 90°