In the figure, O Is the centre of the circle and YC is parallel to ZA. BD = BC, ∠OZY = 57° and ∠DCB = 48°. Find
- ∠AZE
- ∠ZAB
(a)
OY = OZ = Radius
∠YZO = ∠ZYO = 57° (Isosceles triangle, OYB)
∠ZOE
= 57° + 57°
= 114° (Exterior angle of a triangle)
OZ = OE = Radius
∠OEZ
= (180° - 114°) ÷ 2
= 66° ÷ 2
= 33°
∠AZE = 33° (Alternate angles, ZA//YE)
(b)
BD = BC
∠BCD = ∠BDC = 48° (Isosceles triangle BCF)
∠CBD
= 180° - 48° - 48°
= 84°
∠EBA = ∠CBD = 84° (Vertically opposite angles)
∠ZAB
= 180° - 84°
= 96° (Interior angles)
Answer(s): (a) 33°; (b) 96°
