The figure is not drawn to scale. It shows a parallelogram KLMN and an isosceles triangle GKN next to it. ∠JNG = 12°. JH and GF are straight lines. Find
- ∠r
- ∠p + ∠q
- ∠q
(a)
∠r
= 180° - ∠JNG
= 180° - 12°
= 168° (Angles on a straight line)
(b)
∠p = ∠KNM (Parallelogram)
∠p + ∠q
= 180° - 12° - 59°
= 109° (Angles on a straight line)
(c)
∠GKN = ∠p (Corresponding angles)
∠GKN = ∠NGK (Isosceles triangle)
∠p
= 180° - (∠p + ∠q)
= 180° - 109°
= 71°
∠q
= 180° - ∠p - ∠p
= 180° - 71° - 71°
= 38° (Angles sum of triangle)
Answer(s): (a) 168°; (b) 109°; (c) 38°