In the figure, O Is the centre of the circle and HP is parallel to KL. NR = NP, ∠OKH = 55° and ∠RPN = 53°. Find
- ∠LKS
- ∠KLN
(a)
OH = OK = Radius
∠HKO = ∠KHO = 55° (Isosceles triangle, OHB)
∠KOS
= 55° + 55°
= 110° (Exterior angle of a triangle)
OK = OS = Radius
∠OSK
= (180° - 110°) ÷ 2
= 70° ÷ 2
= 35°
∠LKS = 35° (Alternate angles, KL//HE)
(b)
NR = NP
∠NPR = ∠NRP = 53° (Isosceles triangle NPF)
∠PNR
= 180° - 53° - 53°
= 74°
∠SNL = ∠PNR = 74° (Vertically opposite angles)
∠KLN
= 180° - 74°
= 106° (Interior angles)
Answer(s): (a) 35°; (b) 106°
