Answer:
[tex]\boxed{\sin(\theta)=-\frac{7}{58}\sqrt{58}}[/tex]
Step-by-step explanation
The given point [tex](3,-7)[/tex] tells us that the terminal side of [tex]\theta[/tex] is in the fourth quadrant.
From the diagram in the attachment,
We can use the Pythagoras Theorem to find the length of the hypotenuse of the right triangle.
Let the hypotenuse be [tex]h\: units[/tex]. Then,
[tex]h^2=7^2+3^2[/tex]
[tex]h^2=49+9[/tex]
[tex]h^2=58[/tex]
[tex]\Rightarrow h=\sqrt{58}[/tex]
Now we use the sine ratio;
[tex]\sin(\theta)=\frac{Opposite}{Hypotenuse}[/tex].
But since the terminal side of [tex]\theta[/tex] is in the fourth quadrant, the sine ratio must be negative.
This implies that;
[tex]\sin(\theta)=-\frac{7}{\sqrt{58}}[/tex].
We rationalize the denominator to get;
[tex]\sin(\theta)=-\frac{7}{58}\sqrt{58}[/tex].
