In the figure, O Is the centre of the circle and FL is parallel to GH. KN = KL, ∠OGF = 58° and ∠NLK = 52°. Find
- ∠HGP
- ∠GHK
(a)
OF = OG = Radius
∠FGO = ∠GFO = 58° (Isosceles triangle, OFB)
∠GOP
= 58° + 58°
= 116° (Exterior angle of a triangle)
OG = OP = Radius
∠OPG
= (180° - 116°) ÷ 2
= 64° ÷ 2
= 32°
∠HGP = 32° (Alternate angles, GH//FE)
(b)
KN = KL
∠KLN = ∠KNL = 52° (Isosceles triangle KLF)
∠LKN
= 180° - 52° - 52°
= 76°
∠PKH = ∠LKN = 76° (Vertically opposite angles)
∠GHK
= 180° - 76°
= 104° (Interior angles)
Answer(s): (a) 32°; (b) 104°