In the figure, BDF and CDE are straight lines. DB = DC and ED = EF. If ∠CBD = 53°, find
- ∠BDE
- ∠DEF
(a)
∠BDE
= 53° + 53°
= 106° (Exterior angle of a triangle)
(b)
∠EDF
= 180° - 106°
= 74° (Angles on a straight line)
∠DEF
= 180° - 74° - 74°
= 32° (Isosceles triangle)
Answer(s): (a) 106°; (b) 32°