If 6 tan A - 5 =0 , find the value of :
` " " (3 sin A - cos A)/( 5 cos A + 9 sin A) `

Answer

To solve the problem, we need to find the value of the expression \((3 \sin A - \cos A) / (5 \cos A + 9 \sin A)\) given that \(6 \tan A - 5 = 0\).

### Step-by-Step Solution:

1. **Solve for \(\tan A\)**:
\[ 6 \tan A - 5 = 0 \]
Adding 5 to both sides:
\[ 6 \tan A = 5 \]
Dividing both sides by 6:
\[ \tan A = \frac{5}{6} \]

**Hint**: Remember that \(\tan A = \frac{\sin A}{\cos A}\).

2. **Substitute \(\tan A\) into the expression**:
We need to evaluate:
\[ \frac{3 \sin A - \cos A}{5 \cos A + 9 \sin A} \]
We can rewrite \(\sin A\) and \(\cos A\) in terms of \(\tan A\):
\[ \sin A = \tan A \cdot \cos A \]
Therefore,
\[ \sin A = \frac{5}{6} \cos A \]

3. **Substitute \(\sin A\) in the expression**:
Substitute \(\sin A = \frac{5}{6} \cos A\) into the expression:
\[ \frac{3\left(\frac{5}{6} \cos A\right) - \cos A}{5 \cos A + 9\left(\frac{5}{6} \cos A\right)} \]
Simplifying the numerator:
\[ \frac{\frac{15}{6} \cos A - \cos A}{5 \cos A + \frac{45}{6} \cos A} \]
The numerator becomes:
\[ \frac{\frac{15}{6} \cos A - \frac{6}{6} \cos A}{5 \cos A + \frac{45}{6} \cos A} = \frac{\frac{9}{6} \cos A}{5 \cos A + \frac{45}{6} \cos A} \]

4. **Combine terms**:
The numerator simplifies to:
\[ \frac{3}{2} \cos A \]
The denominator simplifies to:
\[ 5 \cos A + \frac{45}{6} \cos A = \frac{30}{6} \cos A + \frac{45}{6} \cos A = \frac{75}{6} \cos A \]

5. **Final expression**:
Now we have:
\[ \frac{\frac{3}{2} \cos A}{\frac{75}{6} \cos A} \]
The \(\cos A\) cancels out:
\[ = \frac{3}{2} \cdot \frac{6}{75} = \frac{18}{150} = \frac{3}{25} \]

### Final Answer:
\[ \frac{3 \sin A - \cos A}{5 \cos A + 9 \sin A} = \frac{3}{25} \]