Question

calcule TG A e TG B em cada triângulo

Answer 1

Resposta:

Explicação passo a passo:

  $tg\:\alpha =\frac{cateto\:\:oposto}{cateto\:adjacente}$

$a)$

   $tg\:\widehat{A}=\frac{BC}{AC}= \frac{4}{3}$

   $tg\:\widehat{B}=\frac{AC}{BC}= \frac{3}{4}$

$b)$

   $(AB)^{2} =(AC)^{2} +(BC)^{2}$

    $(\frac{5\sqrt{2} }{2}) ^{2} =(\frac{5}{2}) ^{2} +(BC)^{2}$

     $\frac{5^{2}.(\sqrt{2}) ^{2} }{2^{2} }=\frac{25}{4} +(BC)^{2}$

     $\frac{25\;.\:2}{4}=\frac{25}{4}+(BC)^{2}$

        $\frac{50}{4}=\frac{25}{4}+(BC)^{2}$

         $\frac{50}{4}-\frac{25}{4}=(BC)^{2}$

         $(BC)^{2}=\frac{25}{4}$

          $BC=\sqrt{\frac{25}{4} }$

          $BC=\frac{5}{2}$

      $tg\:\widehat{A}=\frac{BC}{AC}= \frac{5}{2}:\frac{5}{2}=1$

      $tg\:\widehat{B}=\frac{AC}{BC}= \frac{5}{2}:\frac{5}{2}=1$