Question

Hi thank you so for all your time and help

Answer 1

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:

`\tan (x)=\sqrt[]{3}`

We'll use the second right triangle to show what we want.

First, given a triangle:

We know that the hypotenuse will be:

`\sqrt[]{x^2+y^2}`

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.

`\tan (\alpha)=(y)/(x)`

and

`\tan (\beta)=(x)/(y)`

Now look at the triangle we have:

We can easily check that:

`\sqrt[]{3}=\frac{\sqrt[]{3}}{1}`

So it would be good to find something that relates the ratio between those two numbers: the tangent of 60º!!

Therefore, we have:

`\tan (60^(\circ))=\frac{\sqrt[]{3}}{1}=\sqrt[]{3}`

And the solution to x is:

`x=60^(\circ)`