If $\sin(\angle ACD)=\dfrac{5}{13}$, then the sides of triangle $ACD$ follow a $5$-$12$-$13$ pattern (scaled by some factor). Use $CD=60$ to find that factor.

Because $CD$ is drawn to the hypotenuse, triangle $ACD$ is similar to the large right triangle. Use the relationship $AC^2=AB\cdot AD$.

In Desmos, type k=60/12 to compute the scale factor $k$.

Type a_d=5k and a_c=13k (Desmos will treat these as variables $a_d$ and $a_c$).

Type a_b=a_c^2/a_d and read the value of a_b.

From the diagram, $\angle ADC$ is a right angle, so triangle $ACD$ is a right triangle.

For $\angle ACD$,

So triangle $ACD$ matches the $5$-$12$-$13$ ratio:

Since $CD=12k$ and $CD=60$,

Then

From the diagram, triangle $ABC$ is a right triangle with an altitude from $C$ to the hypotenuse $\overline{AB}$ at point $D$. In this situation, triangle $ACD$ is similar to triangle $ABC$, which leads to the geometric-mean relationship

Substitute $AC=65$ and $AD=25$ into $AC^2=AB\cdot AD$:

Therefore, the length of $\overline{AB}$ is 169.