In the figure, O Is the centre of the circle and EK is parallel to FG. HL = HK, ∠OFE = 52° and ∠LKH = 57°. Find
- ∠GFN
- ∠FGH
(a)
OE = OF = Radius
∠EFO = ∠FEO = 52° (Isosceles triangle, OEB)
∠FON
= 52° + 52°
= 104° (Exterior angle of a triangle)
OF = ON = Radius
∠ONF
= (180° - 104°) ÷ 2
= 76° ÷ 2
= 38°
∠GFN = 38° (Alternate angles, FG//EE)
(b)
HL = HK
∠HKL = ∠HLK = 57° (Isosceles triangle HKF)
∠KHL
= 180° - 57° - 57°
= 66°
∠NHG = ∠KHL = 66° (Vertically opposite angles)
∠FGH
= 180° - 66°
= 114° (Interior angles)
Answer(s): (a) 38°; (b) 114°