To determine the angle that the line given by the equation [tex]\( x \cos \alpha + y \sin \alpha = p \)[/tex] makes with the positive [tex]\( x \)[/tex]-axis, we need to understand the form of the linear equation and how to interpret its parameters in terms of the angle of inclination.
Consider the standard form of a line:
[tex]\[ Ax + By + C = 0 \][/tex]
In this case, the given equation is:
[tex]\[ x \cos \alpha + y \sin \alpha - p = 0 \][/tex]
Here, [tex]\( A = \cos \alpha \)[/tex] and [tex]\( B = \sin \alpha \)[/tex]. Now, the slope [tex]\( m \)[/tex] of the line [tex]\( Ax + By + C = 0 \)[/tex] is given by:
[tex]\[ m = -\frac{A}{B} = -\frac{\cos \alpha}{\sin \alpha} = -\cot \alpha \][/tex]
The angle [tex]\( \theta \)[/tex], which the line makes with the positive [tex]\( x \)[/tex]-axis, is related to the slope by the tangent function:
[tex]\[ m = \tan \theta \][/tex]
Thus, if [tex]\( \tan \theta = -\cot \alpha \)[/tex], then:
[tex]\[ \tan \theta = -\frac{1}{\tan \alpha} = \tan (180^\circ - \alpha) \][/tex]
So, the angle [tex]\( \theta \)[/tex] is:
[tex]\[ \theta = 180^\circ - \alpha \][/tex]
Therefore, the angle made by the line [tex]\( x \cos \alpha + y \sin \alpha = p \)[/tex] with the positive [tex]\( x \)[/tex]-axis is [tex]\( 180^\circ - \alpha \)[/tex].
Hence, the correct option is:
b. [tex]\( 180^\circ - \alpha \)[/tex]
