The figure, not drawn to scale, O is the centre of the circle and KQ // LP // MN. Find
- ∠JOM
- ∠JPO
(a)
∠QOL
= 180° - 96°
= 84° (Interior angles, KQ//LO)
∠OMP = ∠OPM = 28° (Isosceles triangle)
∠LOM
= 28° + 28°
= 56° (Exterior angle of a triangle)
∠JOM
= 84° + 56°
= 140°
(b)
∠POM
= 180° - 28° -28°
= 124° (Isosceles triangle)
∠JOP
= 360° - 140° - 124°
= 96° (Angles at a point)
JO = OP = Radius
∠JPO
= (180° - 96°) ÷ 2
= 84° ÷ 2
= 42° (Isosceles triangle)
Answer(s): (a) 140°; (b) 42°