Return to the Purplemath home page

 The Purplemath Forums
Helping students gain understanding
and self-confidence in algebra


powered by FreeFind

 

Return to the Lessons Index  | Do the Lessons in Order  |  Get "Purplemath on CD" for offline use  |  Print-friendly page

Reducing Angles to Useful Values (page 2 of 2)

Sections: Radians and Degrees, Reducing Angles to Useful Values


For reasons you'll learn more about in trigonometry and calculus, it is generally helpful to have your angles be between 0 and 360. But not all angles will fall within this interval. However, since "once around" takes you right back where you'd started, you can delete revolutions until you get down to an angle between 0 and 360. For instance, an angle of 370 is 10 more than "once around". If you subtract that extra "once around", you'll end up facing in the exact same direction as before, but your angle measure will be the more-manageable 370 360 = 10.

  • Find the angle between 0 and 360 that corresponds to 1275.
  • I can subtract 360's, or I can grab my calculator and do the division: 1275 360 = 3.541666... The only part I care about is the "3", which tells me that 360 fits into 1275 three times:

      1275 3360 = 1275 1080 = 195

    Then my answer is 195.

If you're working "by hand", you can do the long division of 1275 by 360, getting a 3 across the top and a remainder of 195 at the bottom. This gives you the exact same result as the calculator method described above. Copyright Elizabeth Stapel 2010-2011 All Rights Reserved

  • Find the angle between 0 radians and 2π radians that corresponds to  (47pi)/3
  • I need to figure out how many cycles of 2π fit in (47pi)/3:

      (47pi)/3 = (45pi)/3 + (2pi)/3 = 15pi + (2pi)/3 = 14pi + 1pi + (2pi)/3 = 7*2pi + (5pi)/3

    So I see that there are seven times of "one around", with  (5pi)/3  left over.

    The corresponding angle is  (5pi)/3  radians.

  • Find the angle between 0 and 360 that corresponds to 17.
  • A negative angle is one that went around "backwards": instead of rotating the "right" way, they went around the "wrong" way. But I can find the corresponding angle by going back around the "right" way or, which is the same thing for such a small angle, subtracting the negative angle from "once around":

    The corresponding angle is 360 17 = 343

  • Find the angle between 0 radians and 2π radians that corresponds to  -pi/4
  • This one works just like the previous one, but in radians. So I'll work in terms of 2π radians, instead of in terms of 360.

      2pi - pi/4 = (8pi)/4 - pi/4 = (7pi)/4

    The corresponding angle is  (7pi)/4  radians.

  • Find an angle between 0 and 360 that corresponds to 3742.
  • This works somewhat similarly to the previous examples. First I'll find how often 360 fits inside 3742:

      3742 360 = 10.39444...

    But this angle was negative, so I actually need one extra "once around" to carry me into the positive angle values, so I'll use 11 instead of 10:

      3742 + 11 360 = 3742 + 3960 = 218

    The corresponding angle is 218

  • Find the angle between 0 and 360 that corresponds to 15736.
  • This is where a calculator really comes in handy, because this number is just ridiculously large. So I'll do the division:

      15736 360 = 43.711...

    So 360 fits into 15736 forty-three times, with a little left over. This gives me:

      43 360 = 15480

    15736 15480 = 256

    The corresponding angle is 256

<< Previous  Top  |  1 | 2  |  Return to Index

Cite this article as:

Stapel, Elizabeth. "Reducing Angles to Useful Values." Purplemath. Available from
    http://www.purplemath.com/modules/radians2.htm. Accessed
 

 



Purplemath:
  Linking to this site
  Printing pages
  School licensing


Reviews of
Internet Sites:
   Free Help
   Practice
   Et Cetera

The "Homework
   Guidelines"

Study Skills Survey

Tutoring from Purplemath
Find a local math tutor


This lesson may be printed out for your personal use.

Content copyright protected by Copyscape website plagiarism search

 

  Copyright 2010-2012  Elizabeth Stapel   |   About   |   Terms of Use

 

 Feedback   |   Error?