Hello, cluap!

This requires some Trigonometry.

I hope you're prepared.

I'm trying to make an isosceles trapezoid with a base of 26 inches and an altitude of 16 inches.

The sides come in with a 6° (84° ngles).

What I'm trying to figure out is the length of the top section that parallels the base. Code:

B x C
* - - - - - *
/| |\
/ | | \
/6°| | \
/ |16 | \
/ | | \
/ | | \
/ 84° | | \
A * - - - * - - - - - * - - - * D
a E x F a
: - - - - - 26 - - - - - - :

We have isosceles trapezoid $\displaystyle ABCD.$

The base is: .$\displaystyle AD = 26.$

The height is: .$\displaystyle BE = 16$

Let: .$\displaystyle x \:=\:BC \:=\:EF$

Let: .$\displaystyle a \:=\:AE \:=\:FD$

Then: .$\displaystyle a + x + a \:=\:26 \quad\Rightarrow\quad x \:=\:26 - 2a$

. . Hence: .$\displaystyle BC \:=\:x \:=\:26 - 2a$ .[1]

In right triangle $\displaystyle BEA\!:\;\;\tan6^o \:=\:\frac{a}{16} \quad\Rightarrow\quad a \:=\:16\tan6^o \:\approx\:1.68$

Substitute into [1]: .$\displaystyle BC \:=\:26-2(1.68) \;=\;22.64\text{ inches}$