Given the base and angles of an isosceles triangle, how to find length of the two sides? I can't seem to find a textbook solution to this. It is always assumed that the length of the sides is know.

Isolceles triangle
So the base $a$ is known. The bottom angles where $\alpha$ and the two sides $b$ touch, are known. 
What is $h$?
 A: 
solution sketch: 
Note that $h$ divides the triangle into two right triangles (h is the perpendicular bisector (altitude) from the base to the opposite vertex); the angles line $h$ forms with $a$ are right angles. This gives you two right triangles, and you only need one of them to compute the values you need. 
Call the two known (marked) angles $\theta$ (they are equal). 
If you know the length of the side $b$: $\sin\theta = \dfrac{h}{b} \implies h = b\sin\theta$
If $a$ is known, $\tan\theta = \dfrac{h}{a/2} \implies h = \dfrac{a}{2}\tan \theta$. 

Now, using the pythorean theorem to relate your sides, we know that
$$h^2+\left(\frac{a}{2}\right)^2\,=\;b^2\tag{pythogorean theorem}$$
If the length of $b$ is unknown, using the pythagorean theorem, then knowing $a/2$ and $h$ will allow you to solve for $b$.
Knowing $\,b\,$ and $\,h\,$ will allow you to solve for $\,\dfrac{a}{2}\,$ by the pythagorean theorem Then double the value of $a/2$ to get $a$.
A: I have explained it graphically below.

A: You can calculate h 
1) using a and b
(a/2) * (a/2) + h * h = b * b
2) using sine law or cosine law http://www.transtutors.com/math-homework-help/laws-of-triangle/
h/ sinα = b / sin 90
A: from one of the following equations
$$h^2+\left(\frac{a}{2}\right)^2=b^2$$
$$\sin\alpha=\frac{h}{b}$$
$$\cos\alpha=\frac{h/2}{b}=\frac{h}{2b}$$
$$\tan\alpha=\frac{h}{a/2}=\frac{2h}{a}$$
if you know two elements you can find the missing one
