How to Calc Small Base and Sides of Isosceles Trapazoid using Base, Height and Angles I am trying to determine the small base length, and side lengths of an isosceles trapezoid from a big base of $24''$, height of $8''$ and angles a/b of $22.5^\circ$.
I have found all kinds of equations that need 3 sides, or the area, and I have neither.
 A: Looking at this picture

you know $b$ and $h$, and the angle, I'll call $\theta$.  You can, right off the bat, see that
$$\frac{b-a}{2} = c \cos \theta$$
Putting this into the formula
$$h = \sqrt{c^2 - \frac{1}{4} (b-a)^2} = c \sqrt{1-\cos^2\theta}$$
Solving for $c$ gives
$$c = \frac{h}{\sqrt{1-\cos^2\theta}}$$
Now we just need the formula for $a$,
$$a = b - 2 c \cos \theta = b - 2 \frac{h \cos \theta }{\sqrt{1-\cos^2\theta}}$$
Now you just substitute your values ($\theta  =\pi/8, b= 24, h = 8$), and you should have it
A: Suppose your trapacium is $ABCD$ where $AD = BC = c$ since it is an isosceles trapezoid. As a consequence, the $\angle DAB = \angle ABC$ and $\angle ADC = \angle BCD = \alpha$ as depicted in the image:

Suppose $AB = a$ and $CD = b$. Since you know the length of base and height with the size of the one of the angle (sappose they are $h$, $b$, and $\alpha$, respectively), we can find the $a$ and $c$ without using a formula:
Considering the right triangle $AED$,
$$\sin \alpha = \frac{AE}{AD}= \frac{h}{c} \Rightarrow \ c =\frac{h}{\sin \alpha} \tag1$$
$$\tan \alpha = \frac{AE}{DE}= \frac{h}{\frac12 (b-a)} =\frac{2h}{(b-a)} \Rightarrow \ a = b - \frac{2h}{\tan \alpha} \tag2$$
Since you know $h$, $b$, and $\alpha$, you can calculate $c$ and $a$ using the equations $(1)$ and $(2)$, respectively.
Also, it is noteworthy that $DE = CF = \frac{h}{\tan \alpha}$. Note that all of these unknowns are found without using pre-derived equations.
