Answer
0
Work Step by Step
We can find the sine of coterminal angle that belongs to [$0^{\circ}$, $360^{\circ}$) for convenience.
$\sin(1800^{\circ})= \sin(1800^{\circ}-5\cdot360^{\circ})=\sin 0^{\circ}$
The terminal side of $0^{\circ}$ angle passes through (1,0). So $x=1,y=0$ and $r=1$.
$\sin 0^{\circ}=\frac{y}{r}=\frac{0}{1}=0$
