In the figure, O Is the centre of the circle and XB is parallel to YZ. AC = AB, ∠OYX = 59° and ∠CBA = 58°. Find
- ∠ZYD
- ∠YZA
(a)
OX = OY = Radius
∠XYO = ∠YXO = 59° (Isosceles triangle, OXB)
∠YOD
= 59° + 59°
= 118° (Exterior angle of a triangle)
OY = OD = Radius
∠ODY
= (180° - 118°) ÷ 2
= 62° ÷ 2
= 31°
∠ZYD = 31° (Alternate angles, YZ//XE)
(b)
AC = AB
∠ABC = ∠ACB = 58° (Isosceles triangle ABF)
∠BAC
= 180° - 58° - 58°
= 64°
∠DAZ = ∠BAC = 64° (Vertically opposite angles)
∠YZA
= 180° - 64°
= 116° (Interior angles)
Answer(s): (a) 31°; (b) 116°