Let's solve this step-by-step:

Problem:

We are given:

$13\sin A = 5$

Find the value of:

$\frac{5\sin A - 2\cos A}{\tan A}$

Step 1: Solve for $\sin A$ and $\cos A$:

From $13\sin A = 5$, we get:

$\sin A = \frac{5}{13}$

Using the Pythagorean identity:

$\sin^2 A + \cos^2 A = 1$

Substitute $\sin A = \frac{5}{13}$:

$\left(\frac{5}{13}\right)^2 + \cos^2 A = 1$

$\frac{25}{169} + \cos^2 A = 1$

$\cos^2 A = 1 - \frac{25}{169} = \frac{144}{169}$

$\cos A = \pm \frac{12}{13}$

Since the quadrant is not specified, we assume $\cos A > 0$:

$\cos A = \frac{12}{13}$

Step 2: Solve for $\tan A$:

$\tan A = \frac{\sin A}{\cos A}$:

$\tan A = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12}$

Step 3: Evaluate $\frac{5\sin A - 2\cos A}{\tan A}$:

Substitute $\sin A = \frac{5}{13}$, $\cos A = \frac{12}{13}$, and $\tan A = \frac{5}{12}$:

The numerator is:

$5\sin A - 2\cos A = 5\left(\frac{5}{13}\right) - 2\left(\frac{12}{13}\right)$

$= \frac{25}{13} - \frac{24}{13} = \frac{1}{13}$

The denominator is $\tan A = \frac{5}{12}$.

Thus:

$\frac{5\sin A - 2\cos A}{\tan A} = \frac{\frac{1}{13}}{\frac{5}{12}} = \frac{1}{13} \cdot \frac{12}{5} = \frac{12}{65}$

Final Answer:

$\boxed{\frac{12}{65}}$

Let me know if you have any questions or need further explanation!

Related Questions:

1. What happens if $\cos A$ is negative?
2. Can you solve the same problem using a different trigonometric identity?
3. How does $\sin A = \frac{5}{13}$ relate to a right triangle?
4. What is the value of $\cot A$ in this case?
5. Can this process be generalized for any similar expression?

Tip:

Always verify the quadrant of the angle to correctly assign the signs of $\sin$, $\cos$, and $\tan$.