Find the acute angles between the st. Lines :
`y-3x-5=0` and `3y-x+6=0`

Answer

To find the acute angle between the two straight lines given by the equations \( y - 3x - 5 = 0 \) and \( 3y - x + 6 = 0 \), we will follow these steps:

### Step 1: Convert the equations to slope-intercept form
The slope-intercept form of a line is given by \( y = mx + c \), where \( m \) is the slope.

1. For the first line \( y - 3x - 5 = 0 \):
\[ y = 3x + 5 \]
Here, the slope \( m_1 = 3 \).

2. For the second line \( 3y - x + 6 = 0 \):
\[ 3y = x - 6 \implies y = \frac{1}{3}x - 2 \]
Here, the slope \( m_2 = \frac{1}{3} \).

### Step 2: Use the formula for the angle between two lines
The formula for the tangent of the angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is given by:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

### Step 3: Substitute the slopes into the formula
Substituting \( m_1 = 3 \) and \( m_2 = \frac{1}{3} \):
\[ \tan \theta = \left| \frac{3 - \frac{1}{3}}{1 + 3 \cdot \frac{1}{3}} \right| \]

### Step 4: Simplify the expression
1. Calculate \( 3 - \frac{1}{3} \):
\[ 3 = \frac{9}{3} \implies 3 - \frac{1}{3} = \frac{9}{3} - \frac{1}{3} = \frac{8}{3} \]

2. Calculate \( 1 + 3 \cdot \frac{1}{3} \):
\[ 3 \cdot \frac{1}{3} = 1 \implies 1 + 1 = 2 \]

3. Now substitute back into the formula:
\[ \tan \theta = \left| \frac{\frac{8}{3}}{2} \right| = \left| \frac{8}{3} \cdot \frac{1}{2} \right| = \left| \frac{8}{6} \right| = \frac{4}{3} \]

### Step 5: Find the angle \( \theta \)
To find \( \theta \), we take the inverse tangent:
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]

### Step 6: Calculate the angle
Using a calculator, we find:
\[ \theta \approx 53.13^\circ \]

### Conclusion
Thus, the acute angle between the two lines is approximately \( 53^\circ \).

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