In the figure, O Is the centre of the circle and CG is parallel to DE. FH = FG, ∠ODC = 52° and ∠HGF = 53°. Find
- ∠EDK
- ∠DEF
(a)
OC = OD = Radius
∠CDO = ∠DCO = 52° (Isosceles triangle, OCB)
∠DOK
= 52° + 52°
= 104° (Exterior angle of a triangle)
OD = OK = Radius
∠OKD
= (180° - 104°) ÷ 2
= 76° ÷ 2
= 38°
∠EDK = 38° (Alternate angles, DE//CE)
(b)
FH = FG
∠FGH = ∠FHG = 53° (Isosceles triangle FGF)
∠GFH
= 180° - 53° - 53°
= 74°
∠KFE = ∠GFH = 74° (Vertically opposite angles)
∠DEF
= 180° - 74°
= 106° (Interior angles)
Answer(s): (a) 38°; (b) 106°