In the figure, O Is the centre of the circle and BF is parallel to CD. EG = EF, ∠OCB = 53° and ∠GFE = 53°. Find
- ∠DCH
- ∠CDE
(a)
OB = OC = Radius
∠BCO = ∠CBO = 53° (Isosceles triangle, OBB)
∠COH
= 53° + 53°
= 106° (Exterior angle of a triangle)
OC = OH = Radius
∠OHC
= (180° - 106°) ÷ 2
= 74° ÷ 2
= 37°
∠DCH = 37° (Alternate angles, CD//BE)
(b)
EG = EF
∠EFG = ∠EGF = 53° (Isosceles triangle EFF)
∠FEG
= 180° - 53° - 53°
= 74°
∠HED = ∠FEG = 74° (Vertically opposite angles)
∠CDE
= 180° - 74°
= 106° (Interior angles)
Answer(s): (a) 37°; (b) 106°