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=√(a^2+b^2) \\\\\\ c=√((-5)^2+(-6)^2)\implies c=√(61)\impliedby \begin{array}{llll} c\ is\ never\\ negative \end{array} \\\\\\ cos(\theta)=\cfrac{adjacent}{hypotenuse}\implies sin(\theta)=\cfrac{-5}{√(61)} \\\\ -------------------------------\\\\ \textit{now, rationalized} \\\\\\ \cfrac{-5}{√(61)}\cdot \cfrac{√(61)}{√(61)}\implies \cfrac{-5√(61)}{61}`