In the figure, O Is the centre of the circle and CG is parallel to DE. FH = FG, ∠ODC = 59° and ∠HGF = 57°. Find
- ∠EDK
- ∠DEF
(a)
OC = OD = Radius
∠CDO = ∠DCO = 59° (Isosceles triangle, OCB)
∠DOK
= 59° + 59°
= 118° (Exterior angle of a triangle)
OD = OK = Radius
∠OKD
= (180° - 118°) ÷ 2
= 62° ÷ 2
= 31°
∠EDK = 31° (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) 31°; (b) 114°