In the figure, O Is the centre of the circle and ZD is parallel to AB. CE = CD, ∠OAZ = 57° and ∠EDC = 48°. Find
- ∠BAF
- ∠ABC
(a)
OZ = OA = Radius
∠ZAO = ∠AZO = 57° (Isosceles triangle, OZB)
∠AOF
= 57° + 57°
= 114° (Exterior angle of a triangle)
OA = OF = Radius
∠OFA
= (180° - 114°) ÷ 2
= 66° ÷ 2
= 33°
∠BAF = 33° (Alternate angles, AB//ZE)
(b)
CE = CD
∠CDE = ∠CED = 48° (Isosceles triangle CDF)
∠DCE
= 180° - 48° - 48°
= 84°
∠FCB = ∠DCE = 84° (Vertically opposite angles)
∠ABC
= 180° - 84°
= 96° (Interior angles)
Answer(s): (a) 33°; (b) 96°
