In the figure, O Is the centre of the circle and AE is parallel to BC. DF = DE, ∠OBA = 56° and ∠FED = 50°. Find
- ∠CBG
- ∠BCD
(a)
OA = OB = Radius
∠ABO = ∠BAO = 56° (Isosceles triangle, OAB)
∠BOG
= 56° + 56°
= 112° (Exterior angle of a triangle)
OB = OG = Radius
∠OGB
= (180° - 112°) ÷ 2
= 68° ÷ 2
= 34°
∠CBG = 34° (Alternate angles, BC//AE)
(b)
DF = DE
∠DEF = ∠DFE = 50° (Isosceles triangle DEF)
∠EDF
= 180° - 50° - 50°
= 80°
∠GDC = ∠EDF = 80° (Vertically opposite angles)
∠BCD
= 180° - 80°
= 100° (Interior angles)
Answer(s): (a) 34°; (b) 100°