In the figure, O Is the centre of the circle and AE is parallel to BC. DF = DE, ∠OBA = 52° and ∠FED = 51°. Find
- ∠CBG
- ∠BCD
(a)
OA = OB = Radius
∠ABO = ∠BAO = 52° (Isosceles triangle, OAB)
∠BOG
= 52° + 52°
= 104° (Exterior angle of a triangle)
OB = OG = Radius
∠OGB
= (180° - 104°) ÷ 2
= 76° ÷ 2
= 38°
∠CBG = 38° (Alternate angles, BC//AE)
(b)
DF = DE
∠DEF = ∠DFE = 51° (Isosceles triangle DEF)
∠EDF
= 180° - 51° - 51°
= 78°
∠GDC = ∠EDF = 78° (Vertically opposite angles)
∠BCD
= 180° - 78°
= 102° (Interior angles)
Answer(s): (a) 38°; (b) 102°
