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 = 37° (Verticallly opposite angles)
∠RNU
= 37° - 10°
= 27°
∠NQU
= 180° - 105°
= 75° (Angles on a straight line)
∠QUT
= 75° + 27°
= 102° (Exterior angle of a triangle)
(b)
∠NPU
= 102° - 37°
= 65° (Exterior angle of a triangle)
Answer(s): (a) 102°; (b) 65°
