Answer:
[tex]\frac{7\sqrt{58}}{58}[/tex]
Step-by-step explanation:
[tex]\\\text{Given that a point on the terminal side of }\theta \text{ is }(-3,7)\\\\\text{so the terminal side is in Quadrant II}\\\\\text{In the right triangle OMN, using Pythagorean theorem, we get}\\\\\text{ON}=\sqrt{(MN)^2+(OM)^2}\\\\\Rightarrow ON=\sqrt{(7)^2+(-3)^2}\\\\\Rightarrow ON=\sqrt{49+9}\\\\\Rightarrow ON=\sqrt{58}\\[/tex]
[tex]\\\text{we know that in Quadrant II, sine and cosecant are positive.}\\\text{so using the trigonometric ratios, we get}\\\\\sin \theta=\frac{\text{Opposite }}{\text{Hypotenus}}\\\\\Rightarrow \sin \theta=\frac{7}{\sqrt{58}}\\\\\Rightarrow \sin \theta=\frac{7\sqrt{58}}{58}\\[/tex]
