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Definition help on Principal angle 
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#1
Jan205, 10:07 PM

P: 587

Can someone please explain to me what Principal angle is, all i have is a one sentence definition but I dont understand what to do or how to find the answer.
The angle between 00 and 3600 is called the principal angle Find the principal angle of the followings. 5000  1500 12500 1350 I have no clue im new to this. 


#2
Jan205, 10:29 PM

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Between 00 and 3600 ?! They probably meant "between 0° and 360°".
In this case, it helps to visualise the problem by imagining a circle and an arrow or something indicating the angle*. Then they want to know what angle between 0° and 360° does the arrow shows after it has rotated 5000°. Dividing 5000 by 360 tells us how many complete circle the arrow has made. 5000/360 = 13+(125/9). It has made 13 complete circles and 125/9 th of one. Then, (125/9)*360° tells us how many degrees does 125/9 th of a circle corresponds to. It is 320°. *Conventionally, if you draw a cartesian coordinate system through your circle with the origin at the center, 0° correspond to the line given by the x axis and a positive angle increment means rotating counterclockwise. 


#3
Jan405, 01:06 PM

P: 587

Ok I understand sort of, I guess I start at the initial arm in standard position sorry the first question was 500 degrees not 5000 but anyways no big deal. I understand how 5000 degrees made 13 complete circles but the 125/9 is confusing me, I dont know how to get the degrees from that (125/9)*360 degrees gives me 5000 not 320 degrees how do I get 320 degrees and what do I do with this?
So far I went 13 times around the circle from the initial arm how much more do I go? 


#4
Jan405, 01:48 PM

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Definition help on Principal angle
5000°> 'x' rotations Then [tex] x=\frac{5000}{360}=\frac{125}{9}=13\frac{8}{9} rotations [/tex] That fraction of [itex] \frac{8}{9} rotations [/itex] can be computed via the same technique If 360°>one rotation,then 'y'°>8/9 rotations U find 'y'=320°. Follow the same pattern for the other 4 angles. Daniel. 


#5
Jan405, 02:59 PM

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"In this case, it helps to visualise the problem by imagining a circle and an arrow or something indicating the angle*. Then they want to know what angle between 0° and 360° does the arrow shows after it has rotated 5000°. Dividing 5000 by 360 tells us how many complete circle the arrow has made. 5000/360 = 125/9=13+(8/9). It has made 13 complete circles and 8/9 th of one. Then, (8/9)*360° tells us how many degrees does 8/9 th of a circle corresponds to. It is 320°." Sorry for confusing you. 


#6
Jan405, 03:01 PM

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Try it for the angle of 500° and show us what you did if you don't get the right answer. The answer is 140°.



#7
Jan405, 04:01 PM

P: 587

ok I got the answer 140 degrees
doing it the same way as dextercioby, now I am confused on how to sketch this in a circle. I have a cartesian plane going through a circle. For 140 degrees I started with the initial arm in standard position and the terminal arm is in the second quadrant. [tex]x=500/360=25/18=1\frac{7} {18} [/tex] Since it is [tex]1\frac {7} {18} [/tex] then when sketching should I go 1 whole rotation plus the 7/18=140 degrees I know that equals 500 degrees again, taking us back to the main question, but Im trying to understand why did we only look at the fraction and not the 1 whole rotation? If it wasnt one would we still only look at the fraction rotation? Basically now I need some pointers on how to sketch these principal angles, please help thanks wouldnt it have been easier to just do 500 degrees subtract 360? 


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Jan405, 04:11 PM

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#9
Jan405, 04:17 PM

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In general, it you want to find the Principal angle of N°, just subtract N°  360°  360°  ...  360° until what you have left is a number between 0° and 360°. Then that is your principal angle. 


#10
Jan405, 04:23 PM

P: 587

ok so if we had 1000 degrees then we would have to just subtract 360 until we got a number that was between 0 degrees and 360 degrees?
That sounds easy if that's correct, but back to my last post and sketching plz 


#11
Jan405, 04:23 PM

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In case of the negative angles,u must make that rotation of the arm clockwise,while in the case of positive angles,u'll have to make it anticlockwise (trigonometric sense).Any integer number of rotations (no matter the sense) is equivalent to 0° so therefore it's not relevant. For angles in the negative interval (360°,0°),u must find the principal angle,which must be between 0° and 360°.The transformation is achived simply by adding 360°. Example:Consider the angle of 4696°.Divide it by 360°.U'll get: 1 rotation anticlockwise > +360° 1 rotation clockwise>360° x rotations clockwise>4696° [tex] x=\frac{4696}{360}=13\frac{16}{360}[/tex] rotations clockwise. Again,neglect the integer number of rotations (no matter the sense) a,d concentrate upon the fraction. 1 rotation clockwise>360° 16/360 rotations clockwise> y° U find y=16°.However,u need the positive angle correseponding to 16°.And that is [tex] Pos.ang.=16+360=+344 [/tex] ,which is in the interval [0,360°]. Daniel. 


#12
Jan405, 04:36 PM

P: 92

i did some googling, and here's a link that tells a little about principal angles... http://webalgebra.math.uiuc.edu/arbangls/rbai.htm the basic deal is this: if you look at an angle, you're probably interested in some geometrical function of it, as sine, cosine, etc. the "problem" is, that you can't know how you got to that angle if it's less than 360 degrees! perhaps you went right from zero degrees to, say, 140 degrees; but you also could have gone 360+140, or 360+360+140, etc. the bottom line is, that no matter how you got there, the trig functions of ("n" times 360 degrees plus X degrees) are exactly the same as the trig functions of "x degrees" alone! therefore, "x" is refered to as the Principal Angle", and the fun is in finding out what the principal angle is if the "how you got there" is more than 360 degrees. hope that helped, maybe a little... 


#13
Jan405, 04:36 PM

P: 587

ok ok ok Dextercioby thanks soo much lol I really do understand now, and really thanks i get it now totally.
Plusaf thanks for the website and thanks to everyone else who helped me out. 


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