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 = 32° (Isosceles triangle)
∠GOH
= 32° + 32°
= 64° (Exterior angle of a triangle)
∠EOH
= 78° + 64°
= 142°
(b)
∠KOH
= 180° - 32° -32°
= 116° (Isosceles triangle)
∠EOK
= 360° - 142° - 116°
= 102° (Angles at a point)
EO = OK = Radius
∠EKO
= (180° - 102°) ÷ 2
= 78° ÷ 2
= 39° (Isosceles triangle)
Answer(s): (a) 142°; (b) 39°