Let $n$ be an integer, and let $\triangle ABC$ be a right-angles triangle with right angle at $C$. It is given that $\sin A$ and $\sin B$ are the roots of the quadratic equation \[(5n+8)x^2-(7n-20)x+120=0.\] Find the value of $n$
1. Given that $\triangle ABC$ is a right-angled triangle with the right angle at $C$, we know that $\angle A + \angle B = \frac{\pi}{2}$. Therefore, $\sin B = \sin\left(\frac{\pi}{2} - A\right) = \cos A$.
2. Using the Pythagorean identity, we have $\sin^2 A + \cos^2 A = 1$. Since $\sin B = \cos A$, it follows that $\sin^2 A + \sin^2 B = 1$.
3. By Vieta's formulas for the quadratic equation $(5n+8)x^2 - (7n-20)x + 120 = 0$, the sum of the roots $\sin A + \sin B$ is given by:
   \[
   \sin A + \sin B = \frac{7n - 20}{5n + 8}
   \]
   and the product of the roots $\sin A \sin B$ is given by:
   \[
   \sin A \sin B = \frac{120}{5n + 8}
   \]
4. Using the identity $\sin^2 A + \sin^2 B = 1$, we substitute the expressions from Vieta's formulas:
   \[
   1 = (\sin A + \sin B)^2 - 2\sin A \sin B
   \]
   Substituting the values, we get:
   \[
   1 = \left(\frac{7n - 20}{5n + 8}\right)^2 - 2\left(\frac{120}{5n + 8}\right)
   \]
5. Simplify the equation:
   \[
   1 = \frac{(7n - 20)^2 - 240}{(5n + 8)^2}
   \]
   Cross-multiplying to clear the denominator, we get:
   \[
   (5n + 8)^2 = (7n - 20)^2 - 240(5n + 8)
   \]
6. Expanding both sides:
   \[
   (5n + 8)^2 = 25n^2 + 80n + 64
   \]
   \[
   (7n - 20)^2 = 49n^2 - 280n + 400
   \]
   \[
   240(5n + 8) = 1200n + 1920
   \]
7. Substituting these expansions into the equation:
   \[
   25n^2 + 80n + 64 = 49n^2 - 280n + 400 - 1200n - 1920
   \]
   Simplifying the right-hand side:
   \[
   25n^2 + 80n + 64 = 49n^2 - 1480n - 1520
   \]
8. Bringing all terms to one side:
   \[
   25n^2 + 80n + 64 - 49n^2 + 1480n + 1520 = 0
   \]
   \[
   -24n^2 + 1560n + 1584 = 0
   \]
   Dividing by -24:
   \[
   n^2 - 65n - 66 = 0
   \]
9. Solving the quadratic equation:
   \[
   (n - 66)(n + 1) = 0
   \]
   \[
   n = 66 \quad \text{or} \quad n = -1
   \]
10. Since $\sin A + \sin B = \frac{7n - 20}{5n + 8}$ must be positive, we check the values:
    - For $n = 66$:
      \[
      \sin A + \sin B = \frac{7(66) - 20}{5(66) + 8} = \frac{462 - 20}{330 + 8} = \frac{442}{338} > 0
      \]
    - For $n = -1$:
      \[
      \sin A + \sin B = \frac{7(-1) - 20}{5(-1) + 8} = \frac{-7 - 20}{-5 + 8} = \frac{-27}{3} < 0
      \]
    Therefore, $n = -1$ is not valid.

The final answer is $\boxed{66}$