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Given cos(k)= negative square root of 3, which is then divided by 2, and 450 degrees < k < 630 degrees, the smallest possible value for k must be ____ degrees.

 Jan 25, 2022
 #1
avatar+13561 
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the smallest possible value for k must be ____ degrees.

 

Hello Julia!

 

\(cos(k)=-\frac{\sqrt{3}}{2}\\ arccos(-\frac{\sqrt{3}}{2})=k\in\{(150+n\cdot 360)deg,(210+n\cdot 360)deg\}\)

 

\(\{510deg,570deg\} \) \(\subset \{k\) | 450deg < k < 630deg\(\}\)

 

The smallest possible value for k must be: k = 510 degrees.

laugh  !

 Jan 25, 2022
edited by asinus  Jan 25, 2022
edited by asinus  Jan 25, 2022
edited by asinus  Jan 25, 2022
 #2
avatar+117100 
0

 

 

\(cos 30^\circ=-\frac{\sqrt3}{2}\)        (knowledge from a 30,60, 90 triangle)

 

2nd or 3rd quad

 

smallest is   180-30 = 150       next is    180+300=210

next is     150+360= 510   BINGO

 

Just as asinus said!  

 Jan 26, 2022

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