In the figure, BCD is parallel to GHJ and the line GD cuts ∠CGJ into half. Given that DG and CH are straight lines, ∠BCG = 45°, ∠GHF = 32° and ∠CFD = 111°, find
- ∠t
- ∠w
- ∠v
(a)
∠GFH = ∠CFD = 111° (Vertically opposite angles)
∠t
= 180° - 111° - 32°
= 37° (Angles sum of triangle)
(b)
∠w
= 180° - 37°
= 143° (Interior angles)
(c)
∠v
= 180° - 37° - 37° - 32°
= 103° (Angles sum of triangle)
Answer(s): (a) 37°; (b) 143°; (c) 103°