In the figure, O Is the centre of the circle and XB is parallel to YZ. AC = AB, ∠OYX = 56° and ∠CBA = 54°. Find
- ∠ZYD
- ∠YZA
(a)
OX = OY = Radius
∠XYO = ∠YXO = 56° (Isosceles triangle, OXB)
∠YOD
= 56° + 56°
= 112° (Exterior angle of a triangle)
OY = OD = Radius
∠ODY
= (180° - 112°) ÷ 2
= 68° ÷ 2
= 34°
∠ZYD = 34° (Alternate angles, YZ//XE)
(b)
AC = AB
∠ABC = ∠ACB = 54° (Isosceles triangle ABF)
∠BAC
= 180° - 54° - 54°
= 72°
∠DAZ = ∠BAC = 72° (Vertically opposite angles)
∠YZA
= 180° - 72°
= 108° (Interior angles)
Answer(s): (a) 34°; (b) 108°
