In the figure, O Is the centre of the circle and AE is parallel to BC. DF = DE, ∠OBA = 54° and ∠FED = 49°. Find
- ∠CBG
- ∠BCD
(a)
OA = OB = Radius
∠ABO = ∠BAO = 54° (Isosceles triangle, OAB)
∠BOG
= 54° + 54°
= 108° (Exterior angle of a triangle)
OB = OG = Radius
∠OGB
= (180° - 108°) ÷ 2
= 72° ÷ 2
= 36°
∠CBG = 36° (Alternate angles, BC//AE)
(b)
DF = DE
∠DEF = ∠DFE = 49° (Isosceles triangle DEF)
∠EDF
= 180° - 49° - 49°
= 82°
∠GDC = ∠EDF = 82° (Vertically opposite angles)
∠BCD
= 180° - 82°
= 98° (Interior angles)
Answer(s): (a) 36°; (b) 98°
