Respuesta :
[tex]\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}
\qquad
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\\\\
-------------------------------\\\\
cos(t)=-\cfrac{\stackrel{adjacent}{3}}{\stackrel{hypotenuse}{5}}[/tex]
now, the hypotenuse is just a radius unit, so, is never negative, so, the fraction is negative because the numerator is negative, that is, the adjacent side is -3.
now, let's use the pythagorean theorem to find the opposite side.
[tex]\bf \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} \\\\\\ \pm\sqrt{5^2-(-3)^2}=b\implies \pm\sqrt{25-9}=b\implies \pm 4=b[/tex]
ok... so, which is it? the +/-? well, we also know that sin(t) >0, namely that the sine of the angle is positive, so, then is +4 then.
[tex]\bf tan(\theta)=\cfrac{opposite}{adjacent}\qquad \qquad tan(t)=\cfrac{4}{-3}\implies \boxed{tan(t)=-\cfrac{4}{3}}[/tex]
now, the hypotenuse is just a radius unit, so, is never negative, so, the fraction is negative because the numerator is negative, that is, the adjacent side is -3.
now, let's use the pythagorean theorem to find the opposite side.
[tex]\bf \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} \\\\\\ \pm\sqrt{5^2-(-3)^2}=b\implies \pm\sqrt{25-9}=b\implies \pm 4=b[/tex]
ok... so, which is it? the +/-? well, we also know that sin(t) >0, namely that the sine of the angle is positive, so, then is +4 then.
[tex]\bf tan(\theta)=\cfrac{opposite}{adjacent}\qquad \qquad tan(t)=\cfrac{4}{-3}\implies \boxed{tan(t)=-\cfrac{4}{3}}[/tex]
