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 = 47° (Verticallly opposite angles)
∠RNU
= 47° - 10°
= 37°
∠NQU
= 180° - 100°
= 80° (Angles on a straight line)
∠QUT
= 80° + 37°
= 117° (Exterior angle of a triangle)
(b)
∠NPU
= 117° - 47°
= 70° (Exterior angle of a triangle)
Answer(s): (a) 117°; (b) 70°