In the figure, NPQ is parallel to TUV and the line TQ cuts ∠PTV into half. Given that QT and PU are straight lines, ∠NPT = 42°, ∠TUS = 32° and ∠PSQ = 116°, find
- ∠n
- ∠q
- ∠p
(a)
∠TSU = ∠PSQ = 116° (Vertically opposite angles)
∠n
= 180° - 116° - 32°
= 32° (Angles sum of triangle)
(b)
∠q
= 180° - 32°
= 148° (Interior angles)
(c)
∠p
= 180° - 32° - 32° - 32°
= 106° (Angles sum of triangle)
Answer(s): (a) 32°; (b) 148°; (c) 106°