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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 $$\beta =\frac x y$$

## 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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