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Finding angles

  1. Apr 21, 2004 #1
    Find the acute angle that is formed by the line y - (sqrt(3)) x + 1 = 0 and the x-axis.

    better picture here:

    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 :frown:
    Last edited by a moderator: Apr 20, 2017
  2. jcsd
  3. Apr 21, 2004 #2

    matt grime

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    you know the gradient and that's tan of the angle you want
  4. Apr 21, 2004 #3
    so how would I start tackling this question :redface: ?
  5. Apr 21, 2004 #4
    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.
  6. Apr 21, 2004 #5

    matt grime

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


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