In the figure, O Is the centre of the circle and YC is parallel to ZA. BD = BC, ∠OZY = 54° and ∠DCB = 57°. Find
- ∠AZE
- ∠ZAB
(a)
OY = OZ = Radius
∠YZO = ∠ZYO = 54° (Isosceles triangle, OYB)
∠ZOE
= 54° + 54°
= 108° (Exterior angle of a triangle)
OZ = OE = Radius
∠OEZ
= (180° - 108°) ÷ 2
= 72° ÷ 2
= 36°
∠AZE = 36° (Alternate angles, ZA//YE)
(b)
BD = BC
∠BCD = ∠BDC = 57° (Isosceles triangle BCF)
∠CBD
= 180° - 57° - 57°
= 66°
∠EBA = ∠CBD = 66° (Vertically opposite angles)
∠ZAB
= 180° - 66°
= 114° (Interior angles)
Answer(s): (a) 36°; (b) 114°