The figure is not drawn to scale. It shows a parallelogram MNPQ and an isosceles triangle JMQ next to it. ∠LQJ = 12°. LK and JH are straight lines. Find
- ∠t
- ∠r + ∠s
- ∠s
(a)
∠t
= 180° - ∠LQJ
= 180° - 12°
= 168° (Angles on a straight line)
(b)
∠r = ∠MQP (Parallelogram)
∠r + ∠s
= 180° - 12° - 58°
= 110° (Angles on a straight line)
(c)
∠JMQ = ∠r (Corresponding angles)
∠JMQ = ∠QJM (Isosceles triangle)
∠r
= 180° - (∠r + ∠s)
= 180° - 110°
= 70°
∠s
= 180° - ∠r - ∠r
= 180° - 70° - 70°
= 40° (Angles sum of triangle)
Answer(s): (a) 168°; (b) 110°; (c) 40°