Question

#12 with work please

Answer 1

$\bf \textit{let's say, the angle is }\theta \textit{ so then }cos^{-1}\left( \frac{2}{3} \right)=\theta \\\\\\ \textit{this means }cos(\theta )=\cfrac{2}{3}\cfrac{\leftarrow adjacent}{\leftarrow hypotenuse}\impliedby \textit{now, let's find the \underline{opposite}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}$

$\bf \pm\sqrt{3^2-2^2}=b\implies \pm\sqrt{9-4}=b\implies\boxed{ \pm\sqrt{5}=b}\\\\ -------------------------------\\\\ cos^{-1}\left( \frac{2}{3} \right)=\theta \implies sin\left[ cos^{-1}\left( \frac{2}{3} \right) \right]\implies sin(\theta ) \\\\\\ sin(\theta )=\cfrac{\pm\sqrt{5}}{3}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}$

it doesn't say the angle is in a certain quadrant, thus the +/- versions of it are both valid.