The figure is not drawn to scale. It shows a parallelogram MNPQ and an isosceles triangle JMQ next to it. ∠LQJ = 14°. LK and JH are straight lines. Find
- ∠t
- ∠r + ∠s
- ∠s
(a)
∠t
= 180° - ∠LQJ
= 180° - 14°
= 166° (Angles on a straight line)
(b)
∠r = ∠MQP (Parallelogram)
∠r + ∠s
= 180° - 14° - 62°
= 104° (Angles on a straight line)
(c)
∠JMQ = ∠r (Corresponding angles)
∠JMQ = ∠QJM (Isosceles triangle)
∠r
= 180° - (∠r + ∠s)
= 180° - 104°
= 76°
∠s
= 180° - ∠r - ∠r
= 180° - 76° - 76°
= 28° (Angles sum of triangle)
Answer(s): (a) 166°; (b) 104°; (c) 28°