Answer:
Part 1) [tex]MN=x\sqrt{2}\ units[/tex]
Part 2) [tex]LN=x\frac{\sqrt{2}}{2}(1+\sqrt{3})\ units[/tex]
Step-by-step explanation:
The picture of the question in the attached figure
step 1
Find the measure of ang;e M
Remember that the sum of the interior angles in any triangle must be equal to 180 degrees
so
[tex]M+L+N=180^o[/tex]
substitute the given values
[tex]M+45^o+30^o=180^o[/tex]
[tex]M=105^o[/tex]
step 2
Applying the law of sines find the length side MN
[tex]\frac{MN}{sin(45^o)}=\frac{x}{sin(30^o)}[/tex]
Remember that
[tex]sin(45^o)=\frac{\sqrt{2}}{2}[/tex]
[tex]sin(30^o)=\frac{1}{2}[/tex]
substitute
[tex]\frac{MN}{\frac{\sqrt{2}}{2}}=\frac{x}{\frac{1}{2}}[/tex]
[tex]MN=x\sqrt{2}\ units[/tex]
step 3
Find the length side LN
Construct the altitude from M to LN.
In the right triangle of the left
[tex]cos(45^o)=\frac{d_1}{x}[/tex] ---> by CAH (adjacent side divided by the hypotenuse)
Remember that
[tex]cos(45^o)=\frac{\sqrt{2}}{2}[/tex]
substitute
[tex]\frac{\sqrt{2}}{2}=\frac{d_1}{x}[/tex]
[tex]d_1=x\frac{\sqrt{2}}{2}\ units[/tex]
In the right triangle of the right
[tex]tan(30^o)=\frac{d_1}{d_2}[/tex] ---> by TOA (opposite side divided by adjacent side)
Remember that
[tex]tan(30^o)=\frac{\sqrt{3}}{3}[/tex]
[tex]d_1=x\frac{\sqrt{2}}{2}\ units[/tex]
substitute
[tex]\frac{\sqrt{3}}{3}=\frac{x\frac{\sqrt{2}}{2}}{d_2}[/tex]
[tex]d_2=\frac{3x\sqrt{2}}{2\sqrt{3}}\ units[/tex]
simplify
[tex]d_2=\frac{x\sqrt{6}}{2}\ units[/tex]
Find the length side LN
Remember that
[tex]LN=d_1+d_2[/tex]
substitute the values
[tex]LN=(x\frac{\sqrt{2}}{2}+\frac{x\sqrt{6}}{2})\ units[/tex]
simplify
[tex]LN=x\frac{\sqrt{2}}{2}(1+\sqrt{3})\ units[/tex]
