Question

#12 with work please

Answer 1

`\bf \textit{let's say, the angle is }\theta \textit{ so then }cos^(-1)\left( (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√(c^2-a^2)=b\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}`


`\bf \pm√(3^2-2^2)=b\implies \pm√(9-4)=b\implies\boxed{ \pm√(5)=b}\\\\ -------------------------------\\\\ cos^(-1)\left( (2)/(3) \right)=\theta \implies sin\left[ cos^(-1)\left( (2)/(3) \right) \right]\implies sin(\theta ) \\\\\\ sin(\theta )=\cfrac{\pm√(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.