In the figure, ACE and BCD are straight lines. CA = CB and DC = DE. If ∠BAC = 55°, find
- ∠ACD
- ∠CDE
(a)
∠ACD
= 55° + 55°
= 110° (Exterior angle of a triangle)
(b)
∠DCE
= 180° - 110°
= 70° (Angles on a straight line)
∠CDE
= 180° - 70° - 70°
= 40° (Isosceles triangle)
Answer(s): (a) 110°; (b) 40°