#### Answer

$\sin \theta =\dfrac {\sqrt {11}}{6}$

#### Work Step by Step

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