In the figure, O Is the centre of the circle and CG is parallel to DE. FH = FG, ∠ODC = 54° and ∠HGF = 57°. Find
- ∠EDK
- ∠DEF
(a)
OC = OD = Radius
∠CDO = ∠DCO = 54° (Isosceles triangle, OCB)
∠DOK
= 54° + 54°
= 108° (Exterior angle of a triangle)
OD = OK = Radius
∠OKD
= (180° - 108°) ÷ 2
= 72° ÷ 2
= 36°
∠EDK = 36° (Alternate angles, DE//CE)
(b)
FH = FG
∠FGH = ∠FHG = 57° (Isosceles triangle FGF)
∠GFH
= 180° - 57° - 57°
= 66°
∠KFE = ∠GFH = 66° (Vertically opposite angles)
∠DEF
= 180° - 66°
= 114° (Interior angles)
Answer(s): (a) 36°; (b) 114°
