Solving a Trigonometry Problem with Tan A and Tan B

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The discussion focuses on finding the values of Tan A and Tan B for the lines L1 and L2 represented by the equations y=2x and 3y=x-1, respectively. It is established that Tan A equals 2, derived from the slope of line L1. To find Tan B, the second equation needs to be rearranged into standard form to identify its slope. The slope (m) is calculated as Δy/Δx, which corresponds to Tanθ. The conversation emphasizes the importance of recognizing the slope in determining the tangent values for both lines.
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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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