The figure is not drawn to scale. It shows a parallelogram LMNP and an isosceles triangle HLP next to it. ∠KPH = 14°. KJ and HG are straight lines. Find
- ∠s
- ∠q + ∠r
- ∠r
(a)
∠s
= 180° - ∠KPH
= 180° - 14°
= 166° (Angles on a straight line)
(b)
∠q = ∠LPN (Parallelogram)
∠q + ∠r
= 180° - 14° - 55°
= 111° (Angles on a straight line)
(c)
∠HLP = ∠q (Corresponding angles)
∠HLP = ∠PHL (Isosceles triangle)
∠q
= 180° - (∠q + ∠r)
= 180° - 111°
= 69°
∠r
= 180° - ∠q - ∠q
= 180° - 69° - 69°
= 42° (Angles sum of triangle)
Answer(s): (a) 166°; (b) 111°; (c) 42°
