Question

Sendo x um ângulo agudo e sen x=1/3, obter tg x

Answer 1

Explicação passo-a-passo:

$tgx = \frac{senx}{cosx}$

${sen}^{2} x + {cos}^{2} x = 1 \\ {( \frac{1}{3} )}^{2} + {cos}^{2} x = 1 \\ {cos}^{2} x = 1 - \frac{1}{9} \\ {cos}^{2} x = \frac{8}{9} \\ cosx = \sqrt{ \frac{8}{9} } \\ cosx = \frac{2 \sqrt{2} }{3}$

$tgx = \frac{senx}{cosx} \\ tgx = \frac{ \frac{1}{3} }{ \frac{2 \sqrt{2} }{3} } \\ tgx = \frac{1}{3} \times \frac{3}{2 \sqrt{2} } \\ tgx = \frac{3}{6 \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} } \\ tgx = \frac{3 \sqrt{2} }{6 \times 2} \\ tgx = \frac{3 \sqrt{2} }{12} \\ tgx = \frac{ \sqrt{2} }{4}$