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

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Discussion Overview

The discussion revolves around finding the acute angle formed by the line defined by the equation y - (sqrt(3))x + 1 = 0 and the x-axis. The context includes elements of trigonometry and geometry, with participants exploring different methods to approach the problem.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about the relevance of the question to trigonometry.
  • Another participant suggests that the gradient of the line corresponds to the tangent of the angle with the x-axis.
  • A different participant proposes a method involving polar coordinates and simplifies the line equation to y = sqrt(3)x, claiming it makes calculations easier.
  • This participant calculates a point on the line and uses trigonometric functions to find the angle, arriving at t = arctan(3/sqrt(3)), which they convert to 60 degrees or pi/3 radians.
  • Another participant critiques the complexity of the previous explanation but agrees with the conclusion that the gradient represents the tangent of the angle.

Areas of Agreement / Disagreement

Participants generally agree on the relationship between the gradient and the tangent of the angle, but there is disagreement on the complexity of the approach and the necessity of certain steps in the calculation.

Contextual Notes

Some assumptions about the line's behavior and the relevance of polar coordinates are not fully explored. The discussion includes varying levels of mathematical rigor and clarity.

Agent_J
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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 :frown:
 
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you know the gradient and that's tan of the angle you want
 
so how would I start tackling this question :redface: ?
 
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.
 
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.
 

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