The figure is not drawn to scale. It shows a parallelogram KLMN and an isosceles triangle GKN next to it. ∠JNG = 14°. JH and GF are straight lines. Find
- ∠r
- ∠p + ∠q
- ∠q
(a)
∠r
= 180° - ∠JNG
= 180° - 14°
= 166° (Angles on a straight line)
(b)
∠p = ∠KNM (Parallelogram)
∠p + ∠q
= 180° - 14° - 63°
= 103° (Angles on a straight line)
(c)
∠GKN = ∠p (Corresponding angles)
∠GKN = ∠NGK (Isosceles triangle)
∠p
= 180° - (∠p + ∠q)
= 180° - 103°
= 77°
∠q
= 180° - ∠p - ∠p
= 180° - 77° - 77°
= 26° (Angles sum of triangle)
Answer(s): (a) 166°; (b) 103°; (c) 26°
