In the figure, O Is the centre of the circle and BF is parallel to CD. EG = EF, ∠OCB = 55° and ∠GFE = 55°. Find
- ∠DCH
- ∠CDE
(a)
OB = OC = Radius
∠BCO = ∠CBO = 55° (Isosceles triangle, OBB)
∠COH
= 55° + 55°
= 110° (Exterior angle of a triangle)
OC = OH = Radius
∠OHC
= (180° - 110°) ÷ 2
= 70° ÷ 2
= 35°
∠DCH = 35° (Alternate angles, CD//BE)
(b)
EG = EF
∠EFG = ∠EGF = 55° (Isosceles triangle EFF)
∠FEG
= 180° - 55° - 55°
= 70°
∠HED = ∠FEG = 70° (Vertically opposite angles)
∠CDE
= 180° - 70°
= 110° (Interior angles)
Answer(s): (a) 35°; (b) 110°