In the figure, O Is the centre of the circle and BF is parallel to CD. EG = EF, ∠OCB = 57° and ∠GFE = 57°. Find
- ∠DCH
- ∠CDE
(a)
OB = OC = Radius
∠BCO = ∠CBO = 57° (Isosceles triangle, OBB)
∠COH
= 57° + 57°
= 114° (Exterior angle of a triangle)
OC = OH = Radius
∠OHC
= (180° - 114°) ÷ 2
= 66° ÷ 2
= 33°
∠DCH = 33° (Alternate angles, CD//BE)
(b)
EG = EF
∠EFG = ∠EGF = 57° (Isosceles triangle EFF)
∠FEG
= 180° - 57° - 57°
= 66°
∠HED = ∠FEG = 66° (Vertically opposite angles)
∠CDE
= 180° - 66°
= 114° (Interior angles)
Answer(s): (a) 33°; (b) 114°
