Finding Angle of Inclination With a Given Slope

AI Thread Summary
To find the angle of inclination for a line with a slope of -1/2, the arctan function yields -26.565051 degrees, which corresponds to the principal value range of arctan. However, the correct angle of inclination must be adjusted to the second quadrant, resulting in 153 degrees. This adjustment is necessary because the angle of inclination is defined between 0 and π radians. The discussion highlights the importance of understanding the periodic nature of the tangent function and the principal value range of arctan. Properly accounting for these factors leads to the correct determination of the angle of inclination.
onemic
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Homework Statement


Find the angle of inclination of a line when the slope, m, is equal to -1/2

Homework Equations


m = rise/run = tan(φ)

φ=angle of inclination

The Attempt at a Solution



I thought the answer would simply be using arctan on the slope, but my answer gives me -26.565051 degrees when the answer is supposed to be 153 degrees:

arctan(-1/2) = 26.565051
 
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yes, in fact ## 180°-26,6°=153,4° ## it is the same angle... (remember that the period of the tangent is ##\pi##)
 
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Ahhh, thank you! I've been scratching my head at this for a few days now. I don't think the Calculus text I am using(Anton Calculus 6th ed.) ever talked about the period of the tangent being π. Or I somehow missed it.
 
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Remember that the arctan function on a calculator gives answers in the principal value range, ##(-\frac \pi 2,\frac \pi 2)##, but the angle of inclination of a line is between ##0## and ##\pi##. That's why you need the second quadrant angle.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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