When tan theta is -ve why did we assume that it's in the second quad?

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The discussion revolves around the determination of the angle θ between two forces based on the tangent value of -3/4. It is clarified that the angle is in the second quadrant because the principal value of arctan is taken between 0 and π, which excludes the fourth quadrant. The negative sign indicates that the angle is measured in a counter-clockwise direction from the positive x-axis. The confusion regarding the negative sign's association with the x-axis rather than the y-axis is addressed by explaining that the angle is defined as the smaller of the two possible angles formed. Understanding these conventions is crucial for accurately determining the angle's quadrant and its implications.
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The problem says
" Two forces of magnitudes 12 and 15 Newton are applied to a point and the tangesnt of the angle between them is -3/4. Find the resultant of the two forces and the measure of its angle of inclination on the first force.
In the answer it says that the theta belongs to the second quad, and I assume this is because the tan -ve but why didn't it assume it's in the fourth quad? And why doess the -ve sign belong to the 4 on the x-axis not the 3 on the y-axis??
 
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Hi Douna2nd! :smile:
Douna2nd said:
" … the tangesnt of the angle between them is -3/4"

In the answer it says that the theta belongs to the second quad, and I assume this is because the tan -ve but why didn't it assume it's in the fourth quad?

Because the convention is we always take the "principal value" for arctan, and that's between 0 and π (ie, 1st or 2nd quadrant).
And why doess the -ve sign belong to the 4 on the x-axis not the 3 on the y-axis??

Not following you. :confused:
 
Douna2nd said:
And why doess the -ve sign belong to the 4 on the x-axis not the 3 on the y-axis??

tiny-tim said:
Not following you. :confused:

That's because -ve is baby-talk for "negative".
 
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