+82 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.
+14915 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.
!
