Question

necesito ayuda por favor daré corona Utilizando el teorema de Pitágoras encuentre el lado que hace falta o la hipotenusa y luego determine la función trigonométrica indicada en cada caso:

Answer 1

Respuesta:

Explicación paso a paso:

a)

      $12^{2}=4^{2} +b^{2} \\\\144=16 + b^2\\\\144-16=b^2\\\\128=b^2\\\\\sqrt{128}=b$                         $tg=\frac{4}{b} \\\\tg=\frac{4}{\sqrt{128}}\\\\tg=\frac{4}{\sqrt{128}}*\frac{\sqrt{128}}{\sqrt{128}}\\\\tg=\frac{4*\sqrt{128}}{\sqrt{128}*\sqrt{128}}\\\\tg=\frac{4*\sqrt{128}}{(\sqrt{128})^2}$             $\\\\tg=\frac{4\sqrt{128}}{128}\\\\tg=\frac{\sqrt{128}}{32}\\\\tg=\frac{ 8\sqrt{2}}{32}\\\\tg=\frac{\sqrt{2}}{4}$

          $sec=\frac{12}{b}\\\\sec =\frac{12}{\sqrt{128}}\\\\sec=\frac{12}{\sqrt{128}}*\frac{\sqrt{128}}{\sqrt{128}}\\\\sec=\frac{12*\sqrt{128}}{\sqrt{128}*\sqrt{128}}\\\\sec=\frac{12*\sqrt{128}}{(\sqrt{128})^2}$      $sec=\frac{12\sqrt{128}}{128}\\\\sec=\frac{3\sqrt{128}}{32}\\\\sec=\frac{3*(8\sqrt{2})}{32}\\\\sec=\frac{3*8(\sqrt{2})}{32}\\\\sec=\frac{3(\sqrt{2})}{4}$                           $cos=\frac{b}{12}\\\\cos=\frac{\sqrt{128}}{12}\\\\cos=\frac{8\sqrt{2}}{12}\\\\cos=\frac{2\sqrt{2}}{3}$

b)

$h^2=5^2+6^2\\\\h^2=25+36\\\\h^2=61\\\\h=\sqrt{61}$                    $cos=\frac{5}{\sqrt{61}}\\\\cos=\frac{5}{\sqrt{61}}*\frac{\sqrt{61}}{\sqrt{61}}\\\\cos=\frac{5*\sqrt{61}}{ \sqrt{61}*\sqrt{61}}\\\\cos=\frac{5\sqrt{61}}{(\sqrt{61})^2}\\\\cos=\frac{5\sqrt{61}}{61}$              $sen=\frac{5}{\sqrt{61}}\\\\sen=\frac{5}{\sqrt{61}}*\frac{\sqrt{61}}{\sqrt{61}}\\\\sen=\frac{5*\sqrt{61}}{\sqrt{61}*\sqrt{61}}\\\\sen=\frac{5*\sqrt{61}}{(\sqrt{61})^2}\\\\sen=\frac{5\sqrt{61}}{61}$  

$tg = \frac{6}{5}$

c)

$15^2=a^2+4^2\\\\225=a^2+16\\\\225-16=a^2\\\\209=a^2\\\\\sqrt{209}=a$             $csc=\frac{15}{\sqrt{209}}\\\\csc=\frac{15}{\sqrt{209}}*\frac{\sqrt{209}}{\sqrt{209}}\\\\csc=\frac{15*\sqrt{209}}{\sqrt{209}*\sqrt{209}}\\\\csc=\frac{15*\sqrt{209}}{(\sqrt{209})^2}\\\\\\csc=\frac{15\sqrt{209}}{209}$              $sec=\frac{15}{4} \\\\\\\\tg=\frac{\sqrt{209}}{4}$