In the figure, O Is the centre of the circle and HP is parallel to KL. NR = NP, ∠OKH = 59° and ∠RPN = 58°. Find
- ∠LKS
- ∠KLN
(a)
OH = OK = Radius
∠HKO = ∠KHO = 59° (Isosceles triangle, OHB)
∠KOS
= 59° + 59°
= 118° (Exterior angle of a triangle)
OK = OS = Radius
∠OSK
= (180° - 118°) ÷ 2
= 62° ÷ 2
= 31°
∠LKS = 31° (Alternate angles, KL//HE)
(b)
NR = NP
∠NPR = ∠NRP = 58° (Isosceles triangle NPF)
∠PNR
= 180° - 58° - 58°
= 64°
∠SNL = ∠PNR = 64° (Vertically opposite angles)
∠KLN
= 180° - 64°
= 116° (Interior angles)
Answer(s): (a) 31°; (b) 116°