Hello there. To solve this question, we'll have to remember some properties about trigonometric functions and the right triangle.
We want to find x such that:
[tex]\tan (x)=\sqrt[]{3}[/tex]
We'll use the second right triangle to show what we want.
First, given a triangle:
We know that the hypotenuse will be:
[tex]\sqrt[]{x^2+y^2}[/tex]
But most importantly, the tangent of the angles α and β can be calculated by only using the legs of the triangle: it is the ratio between the opposite side and the adjacent side to an angle.
[tex]\tan (\alpha)=\frac{y}{x}[/tex]
and
[tex]\tan (\beta)=\frac{x}{y}[/tex]
Now look at the triangle we have:
We can easily check that:
[tex]\sqrt[]{3}=\frac{\sqrt[]{3}}{1}[/tex]
So it would be good to find something that relates the ratio between those two numbers: the tangent of 60º!!
Therefore, we have:
[tex]\tan (60^{\circ})=\frac{\sqrt[]{3}}{1}=\sqrt[]{3}[/tex]
And the solution to x is:
[tex]x=60^{\circ}[/tex]
