Find the Acute Angle Formed by Line y-(sqrt(3))x+1=0 & x-Axis

[Agent_J]

Find the acute angle that is formed by the line y - (sqrt(3)) x + 1 = 0 and the x-axis.

better picture here:
http://members.rogers.com/agentj/images/math.jpg

I am totally lost with this one. It was from my old trigonometry test, but I don't see the relavance of the question to trigonometry

 

[matt grime]

you know the gradient and that's tan of the angle you want

 

[Agent_J]

so how would I start tackling this question ?

 

[Inspector Gadget]

What this is basically is polar coordinates.

We solve for y, and get y = sqrt(3) - 1. It's easier to do if the graph intersects the origin. Because this is just a straight line, the -1 can be removed from the equation and it won't change the angle to the x-axis. So, our equation is...

y = sqrt(3)x

Next, we need to get a point. Just to keep nice even answers, I'll use x = sqrt(3).

y = sqrt(3)x
y = sqrt(3)*sqrt(3)
y = 3

So, a point is (sqrt(3), 3).

Now, what you do is basically, it makes a triangle. The base (x) is sqrt(3), and the height (y) is 3.

We use our SOH-CAH-TOA trig functions, and see that tan = opp/adj. The opposite angle is the height/y, and the adjacent angle is the base/x.

I'm just using the letter t for now to make things easier...

tan t = 3/sqrt(3)
t = arctan(3/sqrt(3))

t = 60 degrees, or pi/3 if you're working in radians

NOTE: I always seem to goof something up whenever I try to help here, so someone else should just double check what I did.

 

[matt grime]

apart from the fact that you made it far more complicated than it needs to be, that is correct.

y=mx+c

then the gradient is m and that is tan of the angle of the slope, that's all.