Respuesta :

Check the picture below.

[tex]\bf sin(A)=\cfrac{\stackrel{opposite}{8}}{\stackrel{hypotenuse}{17}}~\hfill cos(A)=\cfrac{\stackrel{adjacent}{15}}{\stackrel{hypotenuse}{17}}~\hfill tan(A)=\cfrac{\stackrel{opposite}{8}}{\stackrel{adjacent}{15}} \\\\\\ sin(B)=\cfrac{\stackrel{opposite}{15}}{\stackrel{hypotenuse}{17}}~\hfill cos(B)=\cfrac{\stackrel{adjacent}{8}}{\stackrel{hypotenuse}{17}}~\hfill tan(B)=\cfrac{\stackrel{opposite}{15}}{\stackrel{adjacent}{8}}[/tex]

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