From $\tan \angle BAC=\frac{5}{12}$, write $BC=5k$ and $AC=12k$ for some positive number $k$.

Use $\tan \angle DBC=\frac{DC}{BC}$ to express $DC$ in terms of $k$, then use $AD=AC-DC$.

After finding $BC$ and $DC$, they are the legs of right triangle $DBC$.

Algebra is faster for this problem because the tangent ratios naturally create side ratios. Desmos can verify the scale factor and the final length.

Graph $y=12x-\frac{15}{4}x$ and $y=66$. Click their intersection. Its $x$-coordinate is the scale factor $k$.

Enter $\sqrt{(5\cdot 8)^2+\left(\frac{15\cdot 8}{4}\right)^2}$. This calculates the hypotenuse of triangle $DBC$ from its two legs.

For $\angle BAC$, the opposite side is $BC$ and the adjacent side is $AC$. Since $\tan \angle BAC=\frac{5}{12}$, let

Triangle $DBC$ is also right at $C$. For $\angle DBC$, the opposite side is $DC$ and the adjacent side is $BC$. Therefore,

Since $BC=5k$, it follows that $DC=\frac{15k}{4}$.

Because $D$ lies between $A$ and $C$, $AD=AC-DC$. Thus,

So $BC=40$ and $DC=30$.

In right triangle $DBC$, the legs are $BC=40$ and $DC=30$. Therefore,