Question

PLEASE HELP ASAP!!

Given: sin =-12/13 and tan ß < 0,

Find cos ß and tan ß.

Answer 1

let's keep in mind that the hypotenuse is just a radius unit and thus is always positive, whilst sine and cosine vary per Quadrant.

so we know the tangent is < 0, which is another way to say "tangent is negative", well, that only happens when the cosine and sine differ in sign, and that only happens in the II and IV Quadrants, so the angle β is on either of those Quadrants.

`\sin(\beta )=-\cfrac{12}{13}\implies \stackrel{ \underline{IV~Quadrant} }{\sin(\beta )=\cfrac{\stackrel{opposite}{-12}}{\underset{hypotenuse}{13}}}\hspace{5em}\textit{let's find the \underline{adjacent side}} \\\\\\ \begin{array}{llll} \textit{using the pythagorean theorem} \\\\ a^2+o^2=c^2\implies a=√(c^2 - o^2) \end{array} \qquad \begin{cases} c=\stackrel{hypotenuse}{13}\\ a=adjacent\\ o=\stackrel{opposite}{-12} \end{cases}`

`a=\pm√( 13^2 - (-12)^2)\implies a=\pm√( 169 - 144 ) \implies a=\pm 5\implies \stackrel{ IV~Quadrant }{a=+5} \\\\[-0.35em] ~\dotfill\\\\ \cos(\beta )=\cfrac{\stackrel{adjacent}{5}}{\underset{hypotenuse}{13}}\hspace{5em} \tan(\beta )=\cfrac{\stackrel{opposite}{-12}}{\underset{adjacent}{5}}`