Let `5x+3y=55` and `7x+y=45` be two lines of regression for a bivariate data
The Acute angle between the two lines of regression is

Options

A. `tan^(-1) (19/18)`
B. `tan^(-1) (16/13)`
C. `tan^(-1) (13/16)`
D. `tan^(-1) (8/19)`

Answer

To find the acute angle between the two lines of regression given by the equations \(5x + 3y = 55\) and \(7x + y = 45\), we will follow these steps:

### Step 1: Rewrite the equations in slope-intercept form
We need to express both equations in the form \(y = mx + b\), where \(m\) is the slope.

**For the first equation:**
\[ 5x + 3y = 55 \]
Rearranging gives:
\[ 3y = -5x + 55 \]
Dividing by 3:
\[ y = -\frac{5}{3}x + \frac{55}{3} \]
Thus, the slope \(b_{yx} = -\frac{5}{3}\).

**For the second equation:**
\[ 7x + y = 45 \]
Rearranging gives:
\[ y = -7x + 45 \]
Thus, the slope \(b_{xy} = -7\).

### Step 2: Use the formula for the angle between two lines
The acute angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) can be calculated using the formula:
\[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Here, \(m_1 = -\frac{5}{3}\) and \(m_2 = -7\).

### Step 3: Substitute the slopes into the formula
Substituting the values:
\[ \tan(\theta) = \left| \frac{-\frac{5}{3} - (-7)}{1 + \left(-\frac{5}{3}\right)(-7)} \right| \]
This simplifies to:
\[ \tan(\theta) = \left| \frac{-\frac{5}{3} + 7}{1 + \frac{35}{3}} \right| \]
Calculating the numerator:
\[ -\frac{5}{3} + 7 = -\frac{5}{3} + \frac{21}{3} = \frac{16}{3} \]
Calculating the denominator:
\[ 1 + \frac{35}{3} = \frac{3}{3} + \frac{35}{3} = \frac{38}{3} \]
Thus, we have:
\[ \tan(\theta) = \left| \frac{\frac{16}{3}}{\frac{38}{3}} \right| = \frac{16}{38} = \frac{8}{19} \]

### Step 4: Calculate the angle \(\theta\)
Now, we find \(\theta\):
\[ \theta = \tan^{-1}\left(\frac{8}{19}\right) \]

### Final Answer
The acute angle between the two lines of regression is:
\[ \theta = \tan^{-1}\left(\frac{8}{19}\right) \]