Find Angle between two line `x sinalpha+ycosalpha=p_(1)` and `x sin B +ycosB=p_(2)" "(alpha gt beta)`

Answer

To find the angle between the two lines given by the equations \( x \sin \alpha + y \cos \alpha = p_1 \) and \( x \sin \beta + y \cos \beta = p_2 \), we can follow these steps:

### Step 1: Identify the slopes of the lines
The first line can be rearranged to the slope-intercept form \( y = mx + b \):
\[ y \cos \alpha = p_1 - x \sin \alpha \]
\[ y = -\frac{\sin \alpha}{\cos \alpha} x + \frac{p_1}{\cos \alpha} \]
Thus, the slope \( m_1 \) of the first line is:
\[ m_1 = -\tan \alpha \]

For the second line:
\[ y \cos \beta = p_2 - x \sin \beta \]
\[ y = -\frac{\sin \beta}{\cos \beta} x + \frac{p_2}{\cos \beta} \]
Thus, the slope \( m_2 \) of the second line is:
\[ m_2 = -\tan \beta \]

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

### Step 3: Substitute the slopes into the formula
Substituting \( m_1 \) and \( m_2 \):
\[ \tan \theta = \frac{-\tan \alpha - (-\tan \beta)}{1 + (-\tan \alpha)(-\tan \beta)} \]
This simplifies to:
\[ \tan \theta = \frac{\tan \beta - \tan \alpha}{1 + \tan \alpha \tan \beta} \]

### Step 4: Recognize the tangent subtraction formula
The expression \( \frac{\tan \beta - \tan \alpha}{1 + \tan \alpha \tan \beta} \) is the tangent subtraction formula:
\[ \tan(\beta - \alpha) \]
Thus, we have:
\[ \tan \theta = \tan(\beta - \alpha) \]

### Step 5: Determine the angle \( \theta \)
Since \( \alpha > \beta \), we can express the angle \( \theta \) as:
\[ \theta = \alpha - \beta \]

### Final Answer
The angle between the two lines is:
\[ \theta = \alpha - \beta \]

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