In the figure, O Is the centre of the circle and YC is parallel to ZA. BD = BC, ∠OZY = 60° and ∠DCB = 56°. Find
- ∠AZE
- ∠ZAB
(a)
OY = OZ = Radius
∠YZO = ∠ZYO = 60° (Isosceles triangle, OYB)
∠ZOE
= 60° + 60°
= 120° (Exterior angle of a triangle)
OZ = OE = Radius
∠OEZ
= (180° - 120°) ÷ 2
= 60° ÷ 2
= 30°
∠AZE = 30° (Alternate angles, ZA//YE)
(b)
BD = BC
∠BCD = ∠BDC = 56° (Isosceles triangle BCF)
∠CBD
= 180° - 56° - 56°
= 68°
∠EBA = ∠CBD = 68° (Vertically opposite angles)
∠ZAB
= 180° - 68°
= 112° (Interior angles)
Answer(s): (a) 30°; (b) 112°
