In the figure, O Is the centre of the circle and FL is parallel to GH. KN = KL, ∠OGF = 56° and ∠NLK = 54°. Find
- ∠HGP
- ∠GHK
(a)
OF = OG = Radius
∠FGO = ∠GFO = 56° (Isosceles triangle, OFB)
∠GOP
= 56° + 56°
= 112° (Exterior angle of a triangle)
OG = OP = Radius
∠OPG
= (180° - 112°) ÷ 2
= 68° ÷ 2
= 34°
∠HGP = 34° (Alternate angles, GH//FE)
(b)
KN = KL
∠KLN = ∠KNL = 54° (Isosceles triangle KLF)
∠LKN
= 180° - 54° - 54°
= 72°
∠PKH = ∠LKN = 72° (Vertically opposite angles)
∠GHK
= 180° - 72°
= 108° (Interior angles)
Answer(s): (a) 34°; (b) 108°