Given the square, calculate $\tan{\alpha}.$ Problem: Given the square $ABCD$, let $M$ be the midpoint on the side $|CD|$ and designate $\alpha=\angle AMB$. Calculate $\tan{\alpha}.$
Attempt: We can, without compromising generality, assume that the side of the square is equal to 1. So drawing a figure we get


I know the following: 
1) That $|AM|=|BM|=\sqrt{1^2+\left(\frac{1}{2}\right)^2}=\frac{\sqrt{5}}{2}.$
2) The area of $ABM$ can be expressed in two ways. One way with normal geometry for triangle and another way is by using the areakit involving $\sin{\alpha}$. So: $$\begin{array}{lcl}
A_1 & = & \frac{1\cdot 1}{2} = \frac{1}{2} \\
A_2 & = & \frac{|AM|\cdot|BM|\cdot\sin{\alpha}}{2} = \frac{5}{2}\cdot\sin{\alpha} \\
\end{array}$$
3) I know that $\tan{\alpha}=\frac{\sin{\alpha}}{\cos{\alpha}},$ so finding $\sin{\alpha}$ and $\cos{\alpha}$ and dividing these two will solve this problem.

Setting $A_1=A_2$ yields the equation $$\frac{5}{2}\sin{\alpha}=\frac{1}{2} \ \Longleftrightarrow \ \sin{\alpha} = \frac{1}{5}.$$
Using the law of cosines in the triangle $ABM$ I get 
$$\begin{array}{lcl}
|AB|^2 & = & |AM|^2+|BM|^2 -2|AM||BM|\cos{\alpha} \\
1 & = & \sqrt{5}-\sqrt{5}\cos{\alpha} \ \Leftrightarrow \ \cos{\alpha} = \frac{\sqrt{5}-1}{\sqrt{5}} \\
\end{array}$$
And finally we have $$\tan{\alpha}=\frac{\sin{\alpha}}{\cos{\alpha}}=\frac{\frac{1}{5}}{\frac{\sqrt{5}-1}{\sqrt{5}}} = \frac{5+\sqrt{5}}{20}.$$
But it's not correct. Any idea where I'm making the mistake in my attempt, and is there an easier way of solving this problem?
 A: You have
$$
\tan\frac{\alpha}{2}=\frac{1/2}{1}=\frac{1}{2}
$$
Then use
$$
\tan2\beta=\frac{2\tan\beta}{1-\tan^2\beta}
$$
With $\beta=\alpha/2$, we get
$$
\tan\alpha=\frac{1}{1-1/4}=\frac{4}{3}
$$

Using your tools it can be done as well; set $r=AM$, for simplicity. Then
$$
\frac{1}{2}=\frac{1}{2}r^2\sin\alpha
$$
from the area and
$$
1=r^2+r^2-2r^2\cos\alpha
$$
from the cosine law.
Thus
$$
\sin\alpha=\frac{1}{r^2}\qquad \cos\alpha=\frac{2r^2-1}{2r^2}
$$
Finally
$$
\tan\alpha=\frac{1}{r^2}\frac{2r^2}{2r^2-1}=\frac{2}{2r^2-1}
$$
Since $r=\sqrt{1+\frac{1}{4}}=\sqrt{5}/2$, we have
$$
\tan\alpha=\frac{2}{5/2-1}=\frac{4}{3}
$$
A: You can calculate tan $ \angle AMD$ which equals 2. Also tan $ \angle BMC $ is 2 .
You can take AMD=BMC=$\theta$ and then can easily compute tan ( 180 - $\theta$) = tan $\alpha$ using identity tan (a-b)
Hope it helps and it's easier as you asked . Your solution have certain problems as mentioned in comments. 
A: You've already found the lengths shown below.

Hence
\begin{align}
   u^2 &= \left(\frac{\sqrt 5}{2}\right)^2 -\left(\frac{2}{\sqrt 5}\right)^2 \\
       &= \frac 54 - \frac 45 \\
       &= \frac{9}{20} \\
     u &= \frac{3}{2\sqrt 5}
\end{align}
It follows that
$ \tan \alpha = \dfrac{\frac{2}{\sqrt 5}}{\frac{3}{2\sqrt 5}} = \frac 43$
