In the figure, O Is the centre of the circle and XB is parallel to YZ. AC = AB, ∠OYX = 52° and ∠CBA = 52°. Find
- ∠ZYD
- ∠YZA
(a)
OX = OY = Radius
∠XYO = ∠YXO = 52° (Isosceles triangle, OXB)
∠YOD
= 52° + 52°
= 104° (Exterior angle of a triangle)
OY = OD = Radius
∠ODY
= (180° - 104°) ÷ 2
= 76° ÷ 2
= 38°
∠ZYD = 38° (Alternate angles, YZ//XE)
(b)
AC = AB
∠ABC = ∠ACB = 52° (Isosceles triangle ABF)
∠BAC
= 180° - 52° - 52°
= 76°
∠DAZ = ∠BAC = 76° (Vertically opposite angles)
∠YZA
= 180° - 76°
= 104° (Interior angles)
Answer(s): (a) 38°; (b) 104°