Answer
$sin~162^{\circ} = \frac{\sqrt{5}-1}{4}$
Work Step by Step
$sin~x = sin(180^{\circ}-x)$
We can find the value of $sin~162^{\circ}$:
$sin~162^{\circ} = sin(180^{\circ}-162^{\circ})$
$sin~162^{\circ} = sin(18^{\circ})$
$sin~162^{\circ} = \frac{\sqrt{5}-1}{4}$
