The figure, not drawn to scale, O is the centre of the circle and KQ // LP // MN. Find
- ∠JOM
- ∠JPO
(a)
∠QOL
= 180° - 102°
= 78° (Interior angles, KQ//LO)
∠OMP = ∠OPM = 33° (Isosceles triangle)
∠LOM
= 33° + 33°
= 66° (Exterior angle of a triangle)
∠JOM
= 78° + 66°
= 144°
(b)
∠POM
= 180° - 33° -33°
= 114° (Isosceles triangle)
∠JOP
= 360° - 144° - 114°
= 102° (Angles at a point)
JO = OP = Radius
∠JPO
= (180° - 102°) ÷ 2
= 78° ÷ 2
= 39° (Isosceles triangle)
Answer(s): (a) 144°; (b) 39°