This is a solution from a book involving trusses (in statics), I don't understand how they found the angle $\theta$. What is their method?

This is the given problem:

(the given lengths differ a bit between the problem in the book and the solution manual)

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Extend the horizontals from $A$ and $B$. Note that $AB=6$. The right triangle with hypotenuse $\overline{BD}$ and horizontal/vertical legs has an angle of $30^\circ$ between the hypotenuse and horizontal leg and its horizontal leg is $16$, so its vertical leg is $16\tan30^\circ$. Looking at the comparable triangle with hypotenuse $\overline{AL}$, the vertical leg is $24\tan25^\circ$.
So, the right triangle with angle $\theta$ at $D$ and hypotenuse $\overline{DL}$ has vertical leg $6+16\tan30^\circ-24\tan25^\circ$, so $$\tan\theta=\frac{6+16\tan30^\circ-24\tan25^\circ}{8}.$$