Respostas

  • Usuário do Brainly
2013-01-23T13:12:34-02:00

Observemos que:

 

\text{tg}~60^{\circ}=\sqrt{3}

 

\text{sin}~30^{\circ}=\dfrac{1}{2}

 

\text{cos}~60^{\circ}=\dfrac{1}{2}

 

\text{ctg}~30^{\circ}=\dfrac{3}{\sqrt{3}}

 

Desta maneira, temos:

 

\left({\sqrt{3}-\frac{1}{2}\right)\times\left(\frac{1}{2}-\sqrt{3}\right)

 

Donde, obtemos:

 

\dfrac{2\sqrt{3}-1+12+2\sqrt{3}}{4}

 

\dfrac{4\sqrt{3}+11}{4}\cong4,48