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


Question

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

Options

A) \(30^{\circ}\)

B) \(60^{\circ}\)

C) \(0^{\circ}\)

D) \(45^{\circ}\)

The correct answer is A.

Explanation:

\(\frac 1{\sqrt 3}y-x=0\)
\(\frac y{\sqrt 3}-x=0\)
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 \(\beta =\frac x y\)

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Dicussion (1)

  • \(\frac 1{\sqrt 3}y-x=0\)
    \(\frac y{\sqrt 3}-x=0\)
    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 \(\beta =\frac x y\)

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