In which quadrant is the terminal of an angle in standard position whose measure is $\frac{- 55 π}{3}$ $(-55pi)/3$ ?

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Answer 1 · 2015-02-03T09:14:54

The answer is: $\frac{5}{3} π$ $5/3pi$ .

Every $2 π$ $2pi$ the angles repeat. So it is possible to add, or subtract, at one angle $2 π$ $2pi$ how many times you want.

Let's sum it until we "reach" an angle of "the first round", i.e until the angle is in $[0, 2 π]$ $[0,2pi]$ .

$- \frac{55}{3} π + 2 π = - \frac{55}{3} π + \frac{6}{3} π r = - \frac{49}{3} π$ $-55/3pi+2pi=-55/3pi+6/3pir=-49/3pi$

$- \frac{49}{3} π + \frac{6}{3} π = - \frac{43}{3} π$ $-49/3pi+6/3pi=-43/3pi$

$- \frac{43}{3} π + \frac{6}{3} π = - \frac{37}{3} π$ $-43/3pi+6/3pi=-37/3pi$

$- \frac{37}{3} π + \frac{6}{3} π = - \frac{31}{3} π$ $-37/3pi+6/3pi=-31/3pi$

$- \frac{31}{3} π - \frac{6}{3} π = - \frac{25}{3} π$ $-31/3pi-6/3pi=-25/3pi$

$- \frac{25}{3} π + \frac{6}{3} π = - \frac{19}{3} π$ $-25/3pi+6/3pi=-19/3pi$

$- \frac{19}{3} π + \frac{6}{3} π = - \frac{13}{3} π$ $-19/3pi+6/3pi=-13/3pi$

$- \frac{13}{3} π + \frac{6}{3} π = - \frac{7}{3} π$ $-13/3pi+6/3pi=-7/3pi$

$- \frac{7}{3} π + \frac{6}{3} π = - \frac{π}{3}$ $-7/3pi+6/3pi=-pi/3$

$- \frac{π}{3} + \frac{6}{3} π = \frac{5}{3} π$ $-pi/3+6/3pi=5/3pi$ .

In which quadrant is the terminal of an angle in standard position whose measure is (-55pi)/3−55π3(-55pi)/3?