In the figure, DEFG is a parallelogram. O is the centre of the circle. Find
- ∠DGF
- ∠DEG
(a)
∠OEG = ∠EGO = 28° (Angle properties within a circle)
∠GOE
= 180° - 28° - 28°
= 124° (Angles sum of triangle)
∠ODE + ∠OED = ∠EOG (Exterior angle of a triangle)
∠ODE = ∠OED (Isosceles triangle, OD = OE)
∠ODE
= 124° ÷ 2
= 62° (Exterior angle of a triangle)
∠DGF
= 180° - 62°
= 118° (Interior angles)
(b)
∠DEG
= 180° - 62° - 28°
= 90° (Angle sum of triangles)
Answer(s): (a) 118°; (b) 90°