Answer:
[tex]LJ=15\ units[/tex]
Step-by-step explanation:
see the attached figure to better understand the problem
step 1
Find the length side KJ
In the right triangle JKM
Applying the Pythagoras Theorem
[tex]KJ^{2}=JM^{2}+KM^{2}[/tex]
we have
[tex]JM=3\ units[/tex]
[tex]KM=6\ units[/tex]
substitute
[tex]KJ^{2}=3^{2}+6^{2}[/tex]
[tex]KJ^{2}=45}[/tex]
[tex]KJ=\sqrt{45}\ units[/tex]
simplify
[tex]KJ=3\sqrt{5}\ units[/tex]
step 2
Find the value of cosine of angle MJK in the right triangle JKM
[tex]cos(JKM)=JM/KJ[/tex]
substitute the values
[tex]cos(JKM)=\frac{3}{3\sqrt{5}}[/tex]
simplify
[tex]cos(JKM)=\frac{\sqrt{5}}{5}[/tex] -----> equation A
step 3
Find the value of cosine of angle MJK in the right triangle JKL
[tex]cos(JKM)=KJ/LJ[/tex]
we have
[tex]KJ=3\sqrt{5}\ units[/tex]
[tex]cos(JKM)=\frac{\sqrt{5}}{5}[/tex] ----> remember equation A
substitute the values
[tex]\frac{\sqrt{5}}{5}=\frac{3\sqrt{5}}{LJ}[/tex]
Simplify
[tex]LJ=5(3)=15\ units[/tex]
