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Mathematics 14 Online
OpenStudy (anonymous):

Are there any rules I should keep in mind when using inverse of tangent to find the angle? Because I calculated the angle but I keep getting it wrong, and the calculator is in the correct settings.

OpenStudy (anonymous):

I have these two components: y=0.30N and x=-0.13N, when I calculate the tangent inverse I get -66.57 degrees. But the answer is 113 degrees.

OpenStudy (freckles):

ok so x is neg and y is pos which puts us in the 2nd quadrant

OpenStudy (anonymous):

ok

OpenStudy (freckles):

the thing is is that arctan( ) will only give us angles for 1st and 4th quadrant since its range is only from (-pi/2,pi/2)

OpenStudy (anonymous):

I had forgotten about that. So I need to add to it so that it is in the range?

OpenStudy (freckles):

so you have to add either +180 deg or -180 degree from arctan(y/x)

OpenStudy (anonymous):

ooohhhhkkkkk!!!!!!! that makes so much sense! thank you a lot!!! :)

OpenStudy (anonymous):

the same thing would apply if the x and y are negative right? that's in the third quadrant, so I could add 180 or -180 to get it to the first quadrant.

OpenStudy (freckles):

right

OpenStudy (anonymous):

ok. thanks! I had been thinking waaay to much for this simple thing.

OpenStudy (freckles):

for either 2nd or 3rd quadrant you can do \[\arctan(\frac{y}{x})+(2k+1)180^o\] where k is an integer by the way (2k+1) is odd

OpenStudy (anonymous):

righhttt!! I remember doing stuff like that in calc

OpenStudy (freckles):

cool stuff

OpenStudy (anonymous):

well, thank you again.

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