The figure is not drawn to scale. It shows a parallelogram TUVW and an isosceles triangle QTW next to it. ∠SWQ = 12°. SR and QP are straight lines. Find
- ∠z
- ∠x + ∠y
- ∠y
(a)
∠z
= 180° - ∠SWQ
= 180° - 12°
= 168° (Angles on a straight line)
(b)
∠x = ∠TWV (Parallelogram)
∠x + ∠y
= 180° - 12° - 62°
= 106° (Angles on a straight line)
(c)
∠QTW = ∠x (Corresponding angles)
∠QTW = ∠WQT (Isosceles triangle)
∠x
= 180° - (∠x + ∠y)
= 180° - 106°
= 74°
∠y
= 180° - ∠x - ∠x
= 180° - 74° - 74°
= 32° (Angles sum of triangle)
Answer(s): (a) 168°; (b) 106°; (c) 32°