In the figure, O Is the centre of the circle and BF is parallel to CD. EG = EF, ∠OCB = 59° and ∠GFE = 54°. Find
- ∠DCH
- ∠CDE
(a)
OB = OC = Radius
∠BCO = ∠CBO = 59° (Isosceles triangle, OBB)
∠COH
= 59° + 59°
= 118° (Exterior angle of a triangle)
OC = OH = Radius
∠OHC
= (180° - 118°) ÷ 2
= 62° ÷ 2
= 31°
∠DCH = 31° (Alternate angles, CD//BE)
(b)
EG = EF
∠EFG = ∠EGF = 54° (Isosceles triangle EFF)
∠FEG
= 180° - 54° - 54°
= 72°
∠HED = ∠FEG = 72° (Vertically opposite angles)
∠CDE
= 180° - 72°
= 108° (Interior angles)
Answer(s): (a) 31°; (b) 108°