I've been surprised at how challenging this problem is. Given an isosceles trapezoid, with the larger base b, the four angles, and the two equal sides c know, find the length of the shorter base a. Is the calculation even possible without knowing the height?
[Math] Given three sides of an isosceles trapezoid, find the smaller base side
geometrytrigonometry
Related Solutions
Suppose the longer base length is $b$ and the $2$ angles are $\alpha$ and $\beta$ with height being $h$
The shorter base length is
$$b-h\cot \alpha-h\cot \beta = b-h(\cot \alpha+ \cot \beta)$$
To see this, notice that the height and the base form perpendicular angle.
Assuming that the diagonal sides (I am not sure this is the standard way for calling them in English, so please edit this answer if needed) are $C$ and $D$ and the parallel sides are $A$ and $B$ with $A>B$,
$$ A = \sqrt{C^2-h^2}+\sqrt{D^2-h^2} + B \tag{1}$$
holds by the Pythagorean theorem.
You may just check which permutation of $a,b,c,d$ fulfills this identity.
In $(1)$ I am actually also assuming that the angles on the major base are both acute.
Some signs have to be changed in $(1)$ is this is not the case.
In this case, for instance, we have $$ A = \color{red}{-}\sqrt{C^2-h^2}+\sqrt{D^2-h^2}+B.$$
Best Answer
where $ \alpha = $ angle between the big side and one of equal sides length $c$.