Write an equation for the line with inclination 30 degrees, -intercept 2

Question

I've decided to re-learn my college years. I have text with answers. The problem:

Write an equation for the line with inclination $30$ degrees, $y$-intercept $2$.

Finding the answer with a table of tangent values is easy. But the answer in the back of the book is:

$$y = \frac{\sqrt{3}x}{3} + 2$$

I can graph the book answer and confirm that it's correct. But I have no idea how the author found the answer expressed in fractions and numbers like that.

Answer 1

You may need to backtrack further in your studying if this is stumping you.

But some help:

Draw a circle, call its radius $r=1$. At any point on that circle, the line from the origin to the circle has angle $x$ in radians, and the coordinates of that point are $(r cos(x), r sin(x))$. So your rise over run in the line is expressed as rise of $sin$ and run of $cos$. For $30$ degrees (aka $\frac{\pi}{6}$ radians), these values are $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$, respectively, which gives your slope. Add $2$ to the expression to raise the y-intercept to $2$. Hence, $y=\frac{rise}{run}x+2=\frac{1}{\sqrt{3}}x+2=\frac{\sqrt{3}}{3}x+2$.

Going forward you're going to want to format your posts with Latex so the math reads right, and you will want to tell people exactly what you've done so they know where best to help you.

Answer 2

Another easier way is to draw an equilateral triangle with sides all equal to 2 units.

Divide the figure into two congruent right triangles.

Apply Pythagoras theorem to one of the right triangles to see why $\tan 30^0$ = ${\sqrt 3} \over 3$.