Find the angle in the line \(\frac 1{\sqrt 3}y-x=0\) makes with positive y-axis

\(\frac 1{\sqrt 3}y-x=0\)
\(\frac y{\sqrt 3}-x=0\)
Multiply through with \(\sqrt 3\)
y - x \(\sqrt 3=0\)
y = x \(\sqrt 3\)
Divide through by \(x\)
\(\frac y x=\frac{\sqrt 3} 1\)
but tan \(\theta = \frac y x=\frac{\sqrt 3} 1\)
\(\theta = tan^{-1}(\sqrt 3)= 60^{\circ}\)
The angle 60^{0} is the angle the line makes with the positive x-axis

\(\Theta + \beta = 90\)
\(60 + \beta = 90\)
\(\beta = 90 - 60\)
\(\beta = 30^{\circ}\)
Note that the angle the line \(\frac 1{\sqrt 3}y-x=0\) makes with the positive y-axis is given by tan \(\beta =\frac x y\)