In the figure, BDF and CDE are straight lines. DB = DC and ED = EF. If ∠CBD = 55°, find
- ∠BDE
- ∠DEF
(a)
∠BDE
= 55° + 55°
= 110° (Exterior angle of a triangle)
(b)
∠EDF
= 180° - 110°
= 70° (Angles on a straight line)
∠DEF
= 180° - 70° - 70°
= 40° (Isosceles triangle)
Answer(s): (a) 110°; (b) 40°