In the figure, HKM and JKL are straight lines. KH = KJ and LK = LM. If ∠JHK = 52°, find
- ∠HKL
- ∠KLM
(a)
∠HKL
= 52° + 52°
= 104° (Exterior angle of a triangle)
(b)
∠LKM
= 180° - 104°
= 76° (Angles on a straight line)
∠KLM
= 180° - 76° - 76°
= 28° (Isosceles triangle)
Answer(s): (a) 104°; (b) 28°