In the figure, O Is the centre of the circle and AE is parallel to BC. DF = DE, ∠OBA = 53° and ∠FED = 55°. Find
- ∠CBG
- ∠BCD
(a)
OA = OB = Radius
∠ABO = ∠BAO = 53° (Isosceles triangle, OAB)
∠BOG
= 53° + 53°
= 106° (Exterior angle of a triangle)
OB = OG = Radius
∠OGB
= (180° - 106°) ÷ 2
= 74° ÷ 2
= 37°
∠CBG = 37° (Alternate angles, BC//AE)
(b)
DF = DE
∠DEF = ∠DFE = 55° (Isosceles triangle DEF)
∠EDF
= 180° - 55° - 55°
= 70°
∠GDC = ∠EDF = 70° (Vertically opposite angles)
∠BCD
= 180° - 70°
= 110° (Interior angles)
Answer(s): (a) 37°; (b) 110°