In the figure, O Is the centre of the circle and YC is parallel to ZA. BD = BC, ∠OZY = 59° and ∠DCB = 54°. Find
- ∠AZE
- ∠ZAB
(a)
OY = OZ = Radius
∠YZO = ∠ZYO = 59° (Isosceles triangle, OYB)
∠ZOE
= 59° + 59°
= 118° (Exterior angle of a triangle)
OZ = OE = Radius
∠OEZ
= (180° - 118°) ÷ 2
= 62° ÷ 2
= 31°
∠AZE = 31° (Alternate angles, ZA//YE)
(b)
BD = BC
∠BCD = ∠BDC = 54° (Isosceles triangle BCF)
∠CBD
= 180° - 54° - 54°
= 72°
∠EBA = ∠CBD = 72° (Vertically opposite angles)
∠ZAB
= 180° - 72°
= 108° (Interior angles)
Answer(s): (a) 31°; (b) 108°