Since we know two sides and their adjoining angle, we are able to use the Law of Cosine to solve for the third side.  Recall that the formula for the Law of Cosine is $\displaystyle A^2 + B^2 - 2ABcos(c) = C^2$.  We will begin by defining our variables according to the problem.

We will let 

$\displaystyle A = 6$

$\displaystyle B = 6$

$\displaystyle c = 45$

How do we know that $\displaystyle c$ is the angle we should use?  $\displaystyle c$ is the angle that would be directly opposite of side $\displaystyle C$ (our unknown side) so this is our desired angle to use in the formula.  Now we will plug these values into our formula.

$\displaystyle A^2 + B^2 - 2ABcos(c) = C^2$

$\displaystyle \implies 6^2 + 6^2 - (2)(6)(6)cos(45) = C^2$

$\displaystyle \implies 36 + 36 - 72(\frac{1}{\sqrt{2}}) = C^2$

$\displaystyle \implies 72 - 36\sqrt{2} = C^2$

$\displaystyle \implies 36 + 36 - 72(\frac{1}{\sqrt{2}}) = C^2$$\displaystyle \implies 36 + 36 - 72(\frac{1}{\sqrt{2}}) = C^2$$\displaystyle \implies 21.0883 = C^2$

$\displaystyle \implies 4.59 = C$

Take the triangle below for example:

First, we can figure out the third angle simply because we know all three angles of a triangle must add up to be 180.  So 

$\displaystyle 60 + 60 + \gamma = 180$

$\displaystyle \implies \gamma = 60$

Now we have enough information to use either law.  We will begin by using the law of cosine $\displaystyle A^2 + B^2 - 2ABcos(c) = C^2$.

We will let

$\displaystyle A=7$

$\displaystyle B=7$

$\displaystyle C=60$

Now we will plug these values into our formula

$\displaystyle A^2 + B^2 - 2ABcos(c) = C^2$

$\displaystyle \implies 7^2 + 7^2 - (2)(7)(7)cos(60) = C^2$

$\displaystyle \implies 49 + 49 - 98(\frac{1}{2})= C^2$

$\displaystyle \implies 98 - 49 = C^2$

$\displaystyle \implies 49 = C^2$

$\displaystyle \implies 7 = C^2$

We can also prove that using the Law of Sines will give us the same answer.  Our formula for the Law of Sines is $\displaystyle \frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)} = \frac{C}{sin(\beta)}$.

We will let:

$\displaystyle A=7$

$\displaystyle C=7$

$\displaystyle \alpha = 60$

$\displaystyle \beta = 60$

$\displaystyle \gamma = 60$

We can set up with the relation  $\displaystyle \frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)}$

$\displaystyle \frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)}$

$\displaystyle \implies \frac{7}{sin(60)} = \frac{B}{sin(60)}$

$\displaystyle \implies 7sin(60) = Bsin(60)$     (by cross-multiplication)

$\displaystyle \implies \frac{7sin(60)}{sin(60)} = B$

$\displaystyle \implies 7 = B$       ($\displaystyle sin(60)$ cancels)

Just looking at the formula  $\displaystyle A^2 + B^2 - 2ABcos(c) = C^2$ we can see that this is similar to the Pythagorean Theorem.  We commonly use the Pythagorean Theorem to solve for side $\displaystyle C$, which is what this problem is asking us to do.  Consider the picture below.  We are able to solve for the third side using the information from the picture.  This is the same information given in the question.

Consider the following relationship: $\displaystyle \frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)}$.  Let’s say we know side $\displaystyle A = 9$and $\displaystyle \alpha = 60$.  Plugging this into our relationship we have $\displaystyle \frac{9}{sin(\gamma)} = \frac{B}{sin(60)}$.  We still have two unknowns and no other information to solve for these.  This is not enough information to use the Law of Sines to solve for the unknown sides and angles.

In order to solve this, we need to recall the law of sines.

$\displaystyle \frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

Where $\displaystyle a$, and $\displaystyle b$ are angles, and $\displaystyle A$, and $\displaystyle B$, are opposite side lengths.

Now let's plug in 62 for $\displaystyle a$, 10 for $\displaystyle A$ and 66 for $\displaystyle b$.
Now our equation becomes

$\displaystyle \left(\frac{\sin( 62 )}{ 10 }\right)= \left(\frac{\sin( 66 )}{ b }\right)$

Now we rearrange the equation to solve for $\displaystyle b$

$\displaystyle \\b = \left(\frac{ 10 \sin( 66 )}{\sin( 62 )}\right) \\\\b = -11.3557732263804$

Now we round our answer to the nearest tenth.

$\displaystyle b = -11.36$

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

$\displaystyle b = 11.36$

In order to solve this, we need to recall the law of sines.

$\displaystyle \frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

Where $\displaystyle a$, and $\displaystyle b$ are angles, and $\displaystyle A$, and $\displaystyle B$, are opposite side lengths.

Now let's plug in 41 for $\displaystyle a$, 6 for $\displaystyle A$ and 39 for $\displaystyle b$.
Now our equation becomes

$\displaystyle \left(\frac{\sin( 41 )}{ 6 }\right)= \left(\frac{\sin( 39 )}{ b }\right)$

Now we rearrange the equation to solve for $\displaystyle b$

$\displaystyle \\b = \left(\frac{ 6 \sin( 39 )}{\sin( 41 )}\right) \\\\b = 6.45402256283484$

Now we round our answer to the nearest tenth.

$\displaystyle b = 6.45$

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

In order to solve this, we need to recall the law of sines.

$\displaystyle \frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

Where $\displaystyle a$, and $\displaystyle b$ are angles, and $\displaystyle A$, and $\displaystyle B$, are opposite side lengths.

Now let's plug in 60 for $\displaystyle a$, 3 for $\displaystyle A$ and 51 for $\displaystyle b$.
Now our equation becomes

$\displaystyle \left(\frac{\sin( 60 )}{ 3 }\right)= \left(\frac{\sin( 51 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\displaystyle \\b = \left(\frac{ 3 \sin( 51 )}{\sin( 60 )}\right) \\\\b = 1.57596947141406$

Now we round our answer to the nearest tenth.

$\displaystyle b = 1.58$

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

In order to solve this, we need to recall the law of sines.

$\displaystyle \frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

Where $\displaystyle a$, and $\displaystyle b$ are angles, and $\displaystyle A$, and $\displaystyle B$, are opposite side lengths.

Now let's plug in 45 for $\displaystyle a$, 13 for $\displaystyle A$ and 65 for $\displaystyle b$.
Now our equation becomes

$\displaystyle \left(\frac{\sin( 45 )}{ 13 }\right)= \left(\frac{\sin( 65 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\displaystyle \\b = \left(\frac{ 13 \sin( 65 )}{\sin( 45 )}\right) \\\\b = -15.9863244465377$

Now we round our answer to the nearest tenth.

$\displaystyle b = -15.99$

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

$\displaystyle b = 15.99$

In order to solve this, we need to recall the law of sines.

$\displaystyle \frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

Where $\displaystyle a$, and $\displaystyle b$ are angles, and $\displaystyle A$, and $\displaystyle B$, are opposite side lengths.

Now let's plug in 26 for $\displaystyle a$, 12 for $\displaystyle A$ and 41 for $\displaystyle b$.
Now our equation becomes

$\displaystyle \left(\frac{\sin( 26 )}{ 12 }\right)= \left(\frac{\sin( 41 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\displaystyle \\b = \left(\frac{ 12 \sin( 41 )}{\sin( 26 )}\right) \\\\b = -15.5778376306642$

Now we round our answer to the nearest tenth.

$\displaystyle b = -15.58$

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

$\displaystyle b = 15.58$

In order to solve this, we need to recall the law of sines.

$\displaystyle \frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

Where $\displaystyle a$, and $\displaystyle b$ are angles, and $\displaystyle A$, and $\displaystyle B$, are opposite side lengths.

Now let's plug in 36 for $\displaystyle a$, 6 for $\displaystyle A$ and 58 for $\displaystyle b$.
Now our equation becomes

$\displaystyle \left(\frac{\sin( 36 )}{ 6 }\right)= \left(\frac{\sin( 58 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\displaystyle \\b = \left(\frac{ 6 \sin( 58 )}{\sin( 36 )}\right) \\\\b = -3.69854363983686$

Now we round our answer to the nearest tenth.

$\displaystyle b = -3.7$

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

$\displaystyle b = 3.7$