Question

This is my next Trigonometry question I am needed help on- I just want to be sure I've got it right, or if not, then the steps I need to take to get it right. Thank you.

Answer 1

$\bf \begin{array}{clclll} (-5&,&-6)\\ x&&y\\ a&&b \end{array}\qquad c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \\\\\\ c=\sqrt{(-5)^2+(-6)^2}\implies c=\sqrt{61}\impliedby \begin{array}{llll} c\ is\ never\\ negative \end{array} \\\\\\ cos(\theta)=\cfrac{adjacent}{hypotenuse}\implies sin(\theta)=\cfrac{-5}{\sqrt{61}} \\\\ -------------------------------\\\\ \textit{now, rationalized} \\\\\\ \cfrac{-5}{\sqrt{61}}\cdot \cfrac{\sqrt{61}}{\sqrt{61}}\implies \cfrac{-5\sqrt{61}}{61}$