In the figure, O Is the centre of the circle and ZD is parallel to AB. CE = CD, ∠OAZ = 54° and ∠EDC = 49°. Find
- ∠BAF
- ∠ABC
(a)
OZ = OA = Radius
∠ZAO = ∠AZO = 54° (Isosceles triangle, OZB)
∠AOF
= 54° + 54°
= 108° (Exterior angle of a triangle)
OA = OF = Radius
∠OFA
= (180° - 108°) ÷ 2
= 72° ÷ 2
= 36°
∠BAF = 36° (Alternate angles, AB//ZE)
(b)
CE = CD
∠CDE = ∠CED = 49° (Isosceles triangle CDF)
∠DCE
= 180° - 49° - 49°
= 82°
∠FCB = ∠DCE = 82° (Vertically opposite angles)
∠ABC
= 180° - 82°
= 98° (Interior angles)
Answer(s): (a) 36°; (b) 98°