In the figure, O Is the centre of the circle and YC is parallel to ZA. BD = BC, ∠OZY = 53° and ∠DCB = 53°. Find
- ∠AZE
- ∠ZAB
(a)
OY = OZ = Radius
∠YZO = ∠ZYO = 53° (Isosceles triangle, OYB)
∠ZOE
= 53° + 53°
= 106° (Exterior angle of a triangle)
OZ = OE = Radius
∠OEZ
= (180° - 106°) ÷ 2
= 74° ÷ 2
= 37°
∠AZE = 37° (Alternate angles, ZA//YE)
(b)
BD = BC
∠BCD = ∠BDC = 53° (Isosceles triangle BCF)
∠CBD
= 180° - 53° - 53°
= 74°
∠EBA = ∠CBD = 74° (Vertically opposite angles)
∠ZAB
= 180° - 74°
= 106° (Interior angles)
Answer(s): (a) 37°; (b) 106°