The figure is not drawn to scale. It shows a parallelogram NPQR and an isosceles triangle KNR next to it. ∠MRK = 14°. ML and KJ are straight lines. Find
- ∠u
- ∠s + ∠t
- ∠t
(a)
∠u
= 180° - ∠MRK
= 180° - 14°
= 166° (Angles on a straight line)
(b)
∠s = ∠NRQ (Parallelogram)
∠s + ∠t
= 180° - 14° - 62°
= 104° (Angles on a straight line)
(c)
∠KNR = ∠s (Corresponding angles)
∠KNR = ∠RKN (Isosceles triangle)
∠s
= 180° - (∠s + ∠t)
= 180° - 104°
= 76°
∠t
= 180° - ∠s - ∠s
= 180° - 76° - 76°
= 28° (Angles sum of triangle)
Answer(s): (a) 166°; (b) 104°; (c) 28°