In the figure, BDF and CDE are straight lines. DB = DC and ED = EF. If ∠CBD = 56°, find
- ∠BDE
- ∠DEF
(a)
∠BDE
= 56° + 56°
= 112° (Exterior angle of a triangle)
(b)
∠EDF
= 180° - 112°
= 68° (Angles on a straight line)
∠DEF
= 180° - 68° - 68°
= 44° (Isosceles triangle)
Answer(s): (a) 112°; (b) 44°