The figure, not drawn to scale, O is the centre of the circle and KQ // LP // MN. Find
- ∠JOM
- ∠JPO
(a)
∠QOL
= 180° - 98°
= 82° (Interior angles, KQ//LO)
∠OMP = ∠OPM = 34° (Isosceles triangle)
∠LOM
= 34° + 34°
= 68° (Exterior angle of a triangle)
∠JOM
= 82° + 68°
= 150°
(b)
∠POM
= 180° - 34° -34°
= 112° (Isosceles triangle)
∠JOP
= 360° - 150° - 112°
= 98° (Angles at a point)
JO = OP = Radius
∠JPO
= (180° - 98°) ÷ 2
= 82° ÷ 2
= 41° (Isosceles triangle)
Answer(s): (a) 150°; (b) 41°