The figure, not drawn to scale, O is the centre of the circle and KQ // LP // MN. Find
- ∠JOM
- ∠JPO
(a)
∠QOL
= 180° - 100°
= 80° (Interior angles, KQ//LO)
∠OMP = ∠OPM = 31° (Isosceles triangle)
∠LOM
= 31° + 31°
= 62° (Exterior angle of a triangle)
∠JOM
= 80° + 62°
= 142°
(b)
∠POM
= 180° - 31° -31°
= 118° (Isosceles triangle)
∠JOP
= 360° - 142° - 118°
= 100° (Angles at a point)
JO = OP = Radius
∠JPO
= (180° - 100°) ÷ 2
= 80° ÷ 2
= 40° (Isosceles triangle)
Answer(s): (a) 142°; (b) 40°