In the figure, O Is the centre of the circle and BF is parallel to CD. EG = EF, ∠OCB = 54° and ∠GFE = 58°. Find
- ∠DCH
- ∠CDE
(a)
OB = OC = Radius
∠BCO = ∠CBO = 54° (Isosceles triangle, OBB)
∠COH
= 54° + 54°
= 108° (Exterior angle of a triangle)
OC = OH = Radius
∠OHC
= (180° - 108°) ÷ 2
= 72° ÷ 2
= 36°
∠DCH = 36° (Alternate angles, CD//BE)
(b)
EG = EF
∠EFG = ∠EGF = 58° (Isosceles triangle EFF)
∠FEG
= 180° - 58° - 58°
= 64°
∠HED = ∠FEG = 64° (Vertically opposite angles)
∠CDE
= 180° - 64°
= 116° (Interior angles)
Answer(s): (a) 36°; (b) 116°