AnExtremelyLongName

Guest has unfortunately counted every segment twice O_O

I think I like Melody's method the best because it is a good way to visualize the solution. The official solution is the best if you are familiar with these types of problems.

Wow. I bookmarked it. And yes! I don't think Loki is dead. 

I think it is ten choose four, but I'm not so sure.

BTW when you see a right triangle a thing I always do is imagine it is inscribed in a circle with the hypotenuse as the diameter. That's just what I do, no need to do what I do. But is always nice to do what I do, because doing what I do makes me feel like people exist. I like when people exist, especially people who do what I do.

Anyways, draw a segment from $A$ to to a point on $BC$ in such it way that it is perpendicular to the hypotenuse. We call the point of intersection on the hypotenuse $L$. 

This means $AL\perp{BC}$. 

 - BC = 5 because it is a 3-4-5 triangle.

By AA similarity, we know that triangle $ALC$ is similar to triangle $ABC$. To find AL, we write the following proportion:

$\frac{AL}{3}=\frac{4}{5}$

Solving for AL: 

1. $5AL=12$

2. $AL=\frac{12}{5}$

Now here is how you solve the rest:

1. Consider point $M$, that is the midpoint of $BC$. Because it is the midpoint, we know that $CM=\frac{5}{2}$

2. Use the pythagorean theorem to find $CL$     hint:$\sqrt{AC^2-AL^2}$

3. Now that you found $CL$, we can find $LM$ by evaluating $CM-CL$

4. Since you know $AL$ and $LM$, you can easily find $AM$ by pythagorean theorem.     hint: $\sqrt{AL^2+LM^2}$

Yay

Another way:

Working off of lokiisnotdead, we know that triangle $BCE$ is isosceles. That means angle EBC is 180 - 35 * 2 = 110

 - By supplements we know that FBA is 180-110=70

 - By supplements we know that AFB is 180-102 = 78

Now that you know two angles of triangle AFB, you can find the third angle, which is angle $A$.

     *All angles in a triangle add up to 180