The figure, not drawn to scale, O is the centre of the circle and KQ // LP // MN. Find
- ∠JOM
- ∠JPO
(a)
∠QOL
= 180° - 94°
= 86° (Interior angles, KQ//LO)
∠OMP = ∠OPM = 32° (Isosceles triangle)
∠LOM
= 32° + 32°
= 64° (Exterior angle of a triangle)
∠JOM
= 86° + 64°
= 150°
(b)
∠POM
= 180° - 32° -32°
= 116° (Isosceles triangle)
∠JOP
= 360° - 150° - 116°
= 94° (Angles at a point)
JO = OP = Radius
∠JPO
= (180° - 94°) ÷ 2
= 86° ÷ 2
= 43° (Isosceles triangle)
Answer(s): (a) 150°; (b) 43°