In the figure, O Is the centre of the circle and YC is parallel to ZA. BD = BC, ∠OZY = 55° and ∠DCB = 58°. Find
- ∠AZE
- ∠ZAB
(a)
OY = OZ = Radius
∠YZO = ∠ZYO = 55° (Isosceles triangle, OYB)
∠ZOE
= 55° + 55°
= 110° (Exterior angle of a triangle)
OZ = OE = Radius
∠OEZ
= (180° - 110°) ÷ 2
= 70° ÷ 2
= 35°
∠AZE = 35° (Alternate angles, ZA//YE)
(b)
BD = BC
∠BCD = ∠BDC = 58° (Isosceles triangle BCF)
∠CBD
= 180° - 58° - 58°
= 64°
∠EBA = ∠CBD = 64° (Vertically opposite angles)
∠ZAB
= 180° - 64°
= 116° (Interior angles)
Answer(s): (a) 35°; (b) 116°