Question

What is the value of sin 0 given that (-3,7) is a point on the terminal side 0

Answer 1

Answer:

$\frac{7\sqrt{58}}{58}$

Step-by-step explanation:

$\\\text{Given that a point on the terminal side of }\theta \text{ is }(-3,7)\\\\\text{so the terminal side is in Quadrant II}\\\\\text{In the right triangle OMN, using Pythagorean theorem, we get}\\\\\text{ON}=\sqrt{(MN)^2+(OM)^2}\\\\\Rightarrow ON=\sqrt{(7)^2+(-3)^2}\\\\\Rightarrow ON=\sqrt{49+9}\\\\\Rightarrow ON=\sqrt{58}$

$\\\text{we know that in Quadrant II, sine and cosecant are positive.}\\\text{so using the trigonometric ratios, we get}\\\\\sin \theta=\frac{\text{Opposite }}{\text{Hypotenus}}\\\\\Rightarrow \sin \theta=\frac{7}{\sqrt{58}}\\\\\Rightarrow \sin \theta=\frac{7\sqrt{58}}{58}$