In the figure, BDF and CDE are straight lines. DB = DC and ED = EF. If ∠CBD = 51°, find
- ∠BDE
- ∠DEF
(a)
∠BDE
= 51° + 51°
= 102° (Exterior angle of a triangle)
(b)
∠EDF
= 180° - 102°
= 78° (Angles on a straight line)
∠DEF
= 180° - 78° - 78°
= 24° (Isosceles triangle)
Answer(s): (a) 102°; (b) 24°