In the figure, ACE and BCD are straight lines. CA = CB and DC = DE. If ∠BAC = 53°, find
- ∠ACD
- ∠CDE
(a)
∠ACD
= 53° + 53°
= 106° (Exterior angle of a triangle)
(b)
∠DCE
= 180° - 106°
= 74° (Angles on a straight line)
∠CDE
= 180° - 74° - 74°
= 32° (Isosceles triangle)
Answer(s): (a) 106°; (b) 32°