Question

Please help !! With #17

Answer 1

well, in the II Quadrant, let's recall the sine or opposite side is positive, whilst the cosine or adjacent side is negative, so hmm

`sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\underset{hypotenuse}{13}}\qquad \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies √(c^2 - b^2)=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm√(13^2 - 5^2)=a\implies \pm√(144)=a\implies \pm 12=a\implies \stackrel{II~Quadrant}{-12=a} \\\\[-0.35em] ~\dotfill`

`cos(\theta )=\cfrac{\stackrel{adjacent}{-12}}{\underset{hypotenuse}{13}}~\hfill tan(\theta )=\cfrac{\stackrel{opposite}{5}}{\underset{adjacent}{-12}} ~\hfill cot(\theta )=\cfrac{\stackrel{adjacent}{-12}}{\underset{opposite}{5}}~\hfill \\\\\\ sec(\theta )=\cfrac{\stackrel{hypotenuse}{13}}{\underset{adjacent}{-12}}\hspace{3.2em} csc(\theta )=\cfrac{\stackrel{hypotenuse}{13}}{\underset{opposite}{5}}`