Question

Considere o triângulo equilátero ABC de lado a. 
Sabemos que a altura do triângulo é h=a.(√3/2)
Defina no triângulo AMC o tangente de 30°

( ) tg 30°=√3⁄5
( ) tg 30°=√3⁄3
( ) tg 30°=√3
( ) tg 30°=√3⁄2
( ) tg 30°=3

Answer 1

$tg 30^{o} = \dfrac{cateto.oposto}{cateto.adjacente} = \dfrac{ \frac{a}{2} }{ \frac{a \sqrt{3} }{2} } = \dfrac{a}{2} . \dfrac{2}{a \sqrt{3} } = \dfrac{1}{ \sqrt{3} } = \dfrac{1. \sqrt{3} }{ \sqrt{3}. \sqrt{3} } = \dfrac{ \sqrt{3} }{ 3 }$

Answer 2

$tg \alpha = \ \ \dfrac{medida \ do \ cateto\ oposto}{medida \do \ cateto\ adjacente }$

$tg30 = \dfrac{\dfrac{a}{2}}{a * \dfrac{\sqrt{3}}{2} } \\ \\ \\ tg30 = \dfrac{\dfrac{a}{2}}{ \dfrac{\sqrt{3}*a}{2}} \\ \\ \\ tg30 = \dfrac{a}{2} * \dfrac{2}{ \sqrt{3} *a} \\ \\ \\ tg30 = \dfrac{2a}{2a \sqrt{3}} \\ \\ \\ tg30 = \dfrac{1}{ \sqrt{3} } \\ \\ \\ tg 30 = \dfrac{ 1* \sqrt{3}}{ \sqrt{3}* \sqrt{3}} \\ \\ \\ tg30=\dfrac{\sqrt{3}}{3}$


Resposta:
$\dfrac{ \sqrt{3} }{3}$