Rememeber that:
[tex]cos\theta=\frac{c.a}{h}=\frac{\sqrt[\placeholder{⬚}]{5}}{5}[/tex]
In this case, c.a= square root(5) and h= 5.
using Pythagorean theorem, we will find the opposite leg:
[tex]\begin{gathered} h^2=c.a^2+c.o^2 \\ c.o^2=h^2-c.a^2 \\ c.o=\sqrt[\placeholder{⬚}]{h^2-c.a^2}=\sqrt[\placeholder{⬚}]{5^2-(\sqrt[\placeholder{⬚}]{5})^2}=2\sqrt[\placeholder{⬚}]{5} \end{gathered}[/tex]
The sin of theta is given by:
[tex]sin\theta=\frac{c.o}{h}=\frac{2\sqrt[\placeholder{⬚}]{5}}{5}[/tex]
And now, given that the angle is in the four quadrant.
As you can see the graph, the opposite cateto is negative, therefore:
[tex]sin\theta=\frac{-c.o}{h}=\frac{-2\sqrt[\placeholder{⬚}]{5}}{5}[/tex]
The answer is: C.
