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° - 11°
= 25°
∠NQU
= 180° - 105°
= 75° (Angles on a straight line)
∠QUT
= 75° + 25°
= 100° (Exterior angle of a triangle)
(b)
∠NPU
= 100° - 36°
= 64° (Exterior angle of a triangle)
Answer(s): (a) 100°; (b) 64°