The figure, not drawn to scale, O is the centre of the circle and FL // GK // HJ. Find
- ∠EOH
- ∠EKO
(a)
∠LOG
= 180° - 102°
= 78° (Interior angles, FL//GO)
∠OHK = ∠OKH = 29° (Isosceles triangle)
∠GOH
= 29° + 29°
= 58° (Exterior angle of a triangle)
∠EOH
= 78° + 58°
= 136°
(b)
∠KOH
= 180° - 29° -29°
= 122° (Isosceles triangle)
∠EOK
= 360° - 136° - 122°
= 102° (Angles at a point)
EO = OK = Radius
∠EKO
= (180° - 102°) ÷ 2
= 78° ÷ 2
= 39° (Isosceles triangle)
Answer(s): (a) 136°; (b) 39°