Question

determine os angulos do triangulo de catetos de 6 e 6√3

Answer 1

Bom dia Maria!

Solução!

$Tg\alpha= \dfrac{Cateto~~oposto}{Cateto~~adjacente}\\\\\\\ \alpha=Angulo\\\\\\ \alpha= \dfrac{6}{6 \sqrt{3} }\\\\\\ \alpha= \dfrac{1}{ \sqrt{3} }\\\\\\ \alpha= \dfrac{ \sqrt{3}}{ \sqrt{3} \times \sqrt{3}}\\\\\\ \alpha= \dfrac{ \sqrt{3}}{ \sqrt{9} }\\\\\\ \alpha= \dfrac{ \sqrt{3}}{ 3}\\\\\\ \boxed{\alpha=30\º}$

Virando o triangulo vamos determinar o angulo beta.

$Tg\beta = \dfrac{Cateto~~adjacente}{Cateto~~oposto}\\\\\\ \beta=angulo\\\\\ \beta= \dfrac{6 \sqrt{3} }{6}\\\\\\ \beta= \sqrt{3} \\\\\\ \boxed{\beta=60\º}$

Bom dia!
Bons estudos!