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 = 34° (Isosceles triangle)
∠GOH
= 34° + 34°
= 68° (Exterior angle of a triangle)
∠EOH
= 78° + 68°
= 146°
(b)
∠KOH
= 180° - 34° -34°
= 112° (Isosceles triangle)
∠EOK
= 360° - 146° - 112°
= 102° (Angles at a point)
EO = OK = Radius
∠EKO
= (180° - 102°) ÷ 2
= 78° ÷ 2
= 39° (Isosceles triangle)
Answer(s): (a) 146°; (b) 39°