Solving a Trigonometry Problem with Tan A and Tan B

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SUMMARY

The discussion focuses on calculating the tangent values for two lines, L1 and L2, defined by the equations y=2x and 3y=x-1, respectively. The tangent of angle A (Tan A) for line L1 is definitively established as 2, derived from the slope of the line. For line L2, the participants suggest rearranging the equation into standard form to determine the slope and consequently Tan B. The relationship between the slope and tangent is emphasized, confirming that the slope (m) equals Tanθ.

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Homework Statement


The lines L1 and L2 with equations y=2x and 3y=x-1 respectively,are drawn on the same set of axes. Given that the scales are the same on both axes and that the angle L1 and L2 make with the positive x-asis are A and B respectively,

write down the value of Tan A and the value of Tan B

Homework Equations


Tan=O/A

The Attempt at a Solution



I've figured out tanA which is 2. Why? It doesn't matter what value of x you substitute into L1 you'll always get 2 when you do O which is Y divide by A which is X. I'm using tan=o/a. But, I do not know how to get tan B; it isn't the same thing.
 
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The general (or standard) equation for a straight line is y = mx + c where

m is the slope = Δy/Δx = Tanθ
and c is a constant.

So comparing that with y=2x it's clear that m=2 (and c=0).

I suggest you rearrange the other equation (3y=x-1) into the standard form for a straight line and work out the slope m.
 
Δy/Δx = Tanθ << Ahh you're right. I never noticed even though I used the same method.Silly me. Thanks
 

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