Question

What are expressions for MN and LN? Hint: Construct the altitude from M to LN.

Answer 1

Answer:

Part 1) $MN=x\sqrt{2}\ units$

Part 2) $LN=x\frac{\sqrt{2}}{2}(1+\sqrt{3})\ units$

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

$M+L+N=180^o$

substitute the given values

$M+45^o+30^o=180^o$

$M=105^o$

step 2

Applying the law of sines find the length side MN

$\frac{MN}{sin(45^o)}=\frac{x}{sin(30^o)}$

Remember that

$sin(45^o)=\frac{\sqrt{2}}{2}$

$sin(30^o)=\frac{1}{2}$

substitute

$\frac{MN}{\frac{\sqrt{2}}{2}}=\frac{x}{\frac{1}{2}}$

$MN=x\sqrt{2}\ units$

step 3

Find the length side LN

Construct the altitude from M to LN.

In the right triangle of the left

$cos(45^o)=\frac{d_1}{x}$ ---> by CAH (adjacent side divided by the hypotenuse)

Remember that

$cos(45^o)=\frac{\sqrt{2}}{2}$

substitute

$\frac{\sqrt{2}}{2}=\frac{d_1}{x}$

$d_1=x\frac{\sqrt{2}}{2}\ units$

In the right triangle of the right

$tan(30^o)=\frac{d_1}{d_2}$ ---> by TOA (opposite side divided by adjacent side)

Remember that

$tan(30^o)=\frac{\sqrt{3}}{3}$

$d_1=x\frac{\sqrt{2}}{2}\ units$

substitute

$\frac{\sqrt{3}}{3}=\frac{x\frac{\sqrt{2}}{2}}{d_2}$

$d_2=\frac{3x\sqrt{2}}{2\sqrt{3}}\ units$

simplify

$d_2=\frac{x\sqrt{6}}{2}\ units$

Find the length side LN

Remember that

$LN=d_1+d_2$

substitute the values

$LN=(x\frac{\sqrt{2}}{2}+\frac{x\sqrt{6}}{2})\ units$

simplify

$LN=x\frac{\sqrt{2}}{2}(1+\sqrt{3})\ units$