## Precalculus (6th Edition)

$\sin \theta =\dfrac {\sqrt {11}}{6}$
$\sin \theta =\pm \sqrt {1-\cos ^{2}\theta }=\pm \sqrt {1-\left( \dfrac {5}{6}\right) ^{2}}=\pm \dfrac {\sqrt {11}}{6}$ In quadrant 1 sine is positive so $\sin \theta =\dfrac {\sqrt {11}}{6}$