Question

Please help with this trig question!

Answer 1

we know the cos(α), and we also know the sin(α) is positive, that only happens on the I and II Quadrants, now, our cosine is negative, and our sine is positive, that means, α is in the II Quadrant.

`\cos(\alpha )=\cfrac{\stackrel{adjacent}{-9}}{\underset{hypotenuse}{12}} \qquad \textit{let's find the \underline{opposite side}} \\\\\\ \begin{array}{llll} \textit{using the pythagorean theorem} \\\\ a^2+o^2=c^2\implies o=√(c^2 - a^2) \end{array} \qquad \begin{cases} c=\stackrel{hypotenuse}{12}\\ a=\stackrel{adjacent}{-9}\\ o=\stackrel{opposite}{y} \end{cases}`

`y=\pm√( 12^2 - (-9)^2)\implies y=\pm√( 144 - 81 ) \implies \stackrel{ \textit{II Quadrant} }{y=+√( 63 )} \\\\\\ ~\hfill~\boxed{(-9~~,~~√(63))}~\hfill~`