Answer:
[tex]KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08\\ \\KN=45\sin 50^{\circ}\approx 34.47[/tex]
Step-by-step explanation:
Given:
KL ║ NM ,
LM = 45
m∠M = 50°
KN ⊥ NM
NL ⊥ LM
Find: KN and KL
1. Consider triangle NLM. This is a right triangle, because NL ⊥ LM. In this triangle,
LM = 45
m∠M = 50°
So,
[tex]\tan \angle M=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{NL}{LM}=\dfrac{NL}{45}\\ \\NL=45\tan 50^{\circ}[/tex]
Also
[tex]m\angle LNM=90^{\circ}-50^{\circ}=40^{\circ}[/tex] (angles LNM and M are complementary).
2. Consider triangle NKL. This is a right triangle, because KN ⊥ NM . In this triangle,
[tex]NL=45\tan 50^{\circ}[/tex]
[tex]m\angle KLN=m\angle LNM=40^{\circ}[/tex] (alternate interior angles)
[tex]m\angle KNL=90^{\circ}-40^{\circ}=50^{\circ}[/tex] (angles KNL and KLN are complementary).
So,
[tex]\sin \angle KNL=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{KL}{LN}=\dfrac{KL}{45\tan 50^{\circ}}\\ \\KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08[/tex]
and
[tex]\cos \angle KNL=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{KN}{LN}=\dfrac{KN}{45\tan 50^{\circ}}\\ \\KN=45\tan 50^{\circ}\cos 50^{\circ}=45\sin 50^{\circ}\approx 34.47[/tex]
