document.write( "Question 1112429: what is the reference angle for a 960 degree angle? show all work please \n" ); document.write( "
Algebra.Com's Answer #727473 by KMST(5328)\"\" \"About 
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Here is what a \"960%5Eo\" looks like,
\n" ); document.write( " It's 2 turns and \"240%5Eo\" (all counterclockwise, of course).
\n" ); document.write( "When you divide \"960%5Eo\" by \"360%5Eo\" the quotient is \"2\" and the remainder is \"240%5Eo\" .
\n" ); document.write( "You do not need to divide (unless the angle measure is huge).
\n" ); document.write( "You can keep subtracting \"360%5Eo\" (one turn) until you get less than one turn.
\n" ); document.write( "You can write that as
\n" ); document.write( "\"960%5Eo-360%5Eo-360%5Eo=240%5Eo\" , or \"960%5Eo-2%28360%5Eo%29=240%5Eo\" , or \"960%5Eo-720%5Eo=240%5Eo\" .
\n" ); document.write( "Now you have to work with \"240%5Eo\" , the co-terminal angle of \"960%5Eo\" .
\n" ); document.write( "An angle measuring \"240%5Eo\" is in the third quadrant,
\n" ); document.write( "where all the angles between \"180%5Eo\" and \"270%5Eo\" are located.
\n" ); document.write( "You can write that as
\n" ); document.write( "\"180%5Eo%3C240%5Eo%3C270%5Eo\" .
\n" ); document.write( "The reference angle is a symmetrical angle in quadrant I, and that angle has the same absolute value for all trigonometric functions.
\n" ); document.write( "For quadrant III, it is easy, you just subtract half a turn<, \"180%5Eo\" ,
\n" ); document.write( "and that gives you the angle with the opposite ray for a terminal side:
\n" ); document.write( " The reference angle is \"240%5Eo-180%5Eo=highlight%2860%5Eo%29\" .
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