In the figure, O Is the centre of the circle and XB is parallel to YZ. AC = AB, ∠OYX = 60° and ∠CBA = 51°. Find
- ∠ZYD
- ∠YZA
(a)
OX = OY = Radius
∠XYO = ∠YXO = 60° (Isosceles triangle, OXB)
∠YOD
= 60° + 60°
= 120° (Exterior angle of a triangle)
OY = OD = Radius
∠ODY
= (180° - 120°) ÷ 2
= 60° ÷ 2
= 30°
∠ZYD = 30° (Alternate angles, YZ//XE)
(b)
AC = AB
∠ABC = ∠ACB = 51° (Isosceles triangle ABF)
∠BAC
= 180° - 51° - 51°
= 78°
∠DAZ = ∠BAC = 78° (Vertically opposite angles)
∠YZA
= 180° - 78°
= 102° (Interior angles)
Answer(s): (a) 30°; (b) 102°