Chapter 6 - Review Exercises - Page 285: 14

$y=60^o$

RECALL: $y=\arccos{(x)} \longrightarrow \cos{y}=x$, $y$ is in the interval $[0^o, 90^o]$ Thus, $y=\arccos{(\frac{1}{2})}$ implies that $\cos{y}=\frac{1}{2}$. Note that $\cos{(60^o)}=\frac{1}{2}$ Therefore, $y=\arccos{(\frac{1}{2})}\longrightarrow y=60^o$