In the figure, HKM and JKL are straight lines. KH = KJ and LK = LM. If ∠JHK = 51°, find
- ∠HKL
- ∠KLM
(a)
∠HKL
= 51° + 51°
= 102° (Exterior angle of a triangle)
(b)
∠LKM
= 180° - 102°
= 78° (Angles on a straight line)
∠KLM
= 180° - 78° - 78°
= 24° (Isosceles triangle)
Answer(s): (a) 102°; (b) 24°