In the figure, O Is the centre of the circle and ZD is parallel to AB. CE = CD, ∠OAZ = 59° and ∠EDC = 55°. Find
- ∠BAF
- ∠ABC
(a)
OZ = OA = Radius
∠ZAO = ∠AZO = 59° (Isosceles triangle, OZB)
∠AOF
= 59° + 59°
= 118° (Exterior angle of a triangle)
OA = OF = Radius
∠OFA
= (180° - 118°) ÷ 2
= 62° ÷ 2
= 31°
∠BAF = 31° (Alternate angles, AB//ZE)
(b)
CE = CD
∠CDE = ∠CED = 55° (Isosceles triangle CDF)
∠DCE
= 180° - 55° - 55°
= 70°
∠FCB = ∠DCE = 70° (Vertically opposite angles)
∠ABC
= 180° - 70°
= 110° (Interior angles)
Answer(s): (a) 31°; (b) 110°