In the figure, O Is the centre of the circle and ZD is parallel to AB. CE = CD, ∠OAZ = 56° and ∠EDC = 54°. Find
- ∠BAF
- ∠ABC
(a)
OZ = OA = Radius
∠ZAO = ∠AZO = 56° (Isosceles triangle, OZB)
∠AOF
= 56° + 56°
= 112° (Exterior angle of a triangle)
OA = OF = Radius
∠OFA
= (180° - 112°) ÷ 2
= 68° ÷ 2
= 34°
∠BAF = 34° (Alternate angles, AB//ZE)
(b)
CE = CD
∠CDE = ∠CED = 54° (Isosceles triangle CDF)
∠DCE
= 180° - 54° - 54°
= 72°
∠FCB = ∠DCE = 72° (Vertically opposite angles)
∠ABC
= 180° - 72°
= 108° (Interior angles)
Answer(s): (a) 34°; (b) 108°