Given the following value
[tex]\cos\theta=-\frac{3}{7}[/tex]We want to calculate
[tex]\sin\theta[/tex]To calculate the sine of the angle, we can use the following identity:
[tex]\cos^2\theta+\sin^2\theta=1[/tex]If we substitute the given cosine value on this identity, we have:
[tex](-\frac{3}{7})^2+\sin^2\theta=1[/tex]Solving for the sine:
[tex]\begin{gathered} (-\frac{3}{7})^2+\sin^2\theta=1 \\ \frac{9}{49}+\sin^2\theta=1 \\ \sin^2\theta=1-\frac{9}{49} \\ \sin^2\theta=\frac{49}{49}-\frac{9}{49} \\ \sin^2\theta=\frac{40}{49} \\ \sin\theta=\pm\sqrt{\frac{40}{49}} \\ \sin\theta=\pm\frac{2\sqrt{10}}{7} \end{gathered}[/tex]Since the angle belongs to the quadrant II, the value of the sine is positive, therefoer, the sine of our angle is:
[tex]\sin\theta=\frac{2\sqrt{10}}{7}[/tex]