In the figure, JKL and LMN are isosceles triangles where JL = KL and LM = LN. Given that MP, NR and JT are straight lines, find
- ∠QUT
- ∠NPU
(a)
∠JKL = ∠KJL = ∠LMN = ∠LNM (Isosceles triangle)
∠LNM = ∠PNU = 36° (Verticallly opposite angles)
∠RNU
= 36° - 12°
= 24°
∠NQU
= 180° - 98°
= 82° (Angles on a straight line)
∠QUT
= 82° + 24°
= 106° (Exterior angle of a triangle)
(b)
∠NPU
= 106° - 36°
= 70° (Exterior angle of a triangle)
Answer(s): (a) 106°; (b) 70°