Rysunek poglądowy:

 

$\mathrm{tg}{30}^o=\frac{\left|CD\right|}{\left|AD\right|}$    

$\left|AD\right|=\frac{\left|CD\right|}{\mathrm{tg}{30}^o}$  

$\left|AD\right|=\frac{3}{\frac{\sqrt{3}}{3}}$  

$\left|AD\right|=3\cdot \frac{3}{\sqrt{3}}=\frac{1}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3}$  

 

$\sin {30}^o=\frac{\left|CD\right|}{\left|AC\right|}$  

$\frac{1}{2}=\frac{3}{\left|AC\right|}$  

$\left|AC\right|=6$  

 

$\mathrm{tg}{60}^o=\frac{\left|CD\right|}{\left|BD\right|}$  

$\left|BD\right|=\frac{\left|CD\right|}{\mathrm{tg}{60}^o}$  

$\left|BD\right|=\frac{3}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{3\sqrt{3}}{3}=\sqrt{3}$  

 

$\sin {60}^o=\frac{\left|CD\right|}{\left|BC\right|}$  

$\left|BC\right|=\frac{\left|CD\right|}{\sin {60}^o}$  

$\left|BC\right|=\frac{3}{\frac{\sqrt{3}}{2}}$  

$\left|BC\right|=3\cdot \frac{2}{\sqrt{3}}=\frac{6}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{6\sqrt{3}}{3}=2\sqrt{3}$  

 

A więc:

$\left|BC\right|=2\sqrt{3}$  

$\left|AC\right|=6$  

$\left|AB\right|=\left|BD\right|+\left|AD\right|=\sqrt{3}+\frac{\sqrt{3}}{3}=\frac{3\sqrt{3}}{3}+\frac{\sqrt{3}}{3}=\frac{4\sqrt{3}}{3}$  

 

Obwód:

$\left|BC\right|+\left|AC\right|+\left|AB\right|=2\sqrt{3}+6+\frac{4\sqrt{3}}{3}=\frac{6\sqrt{3}}{3}+\frac{18}{3}+\frac{4\sqrt{3}}{3}=\frac{10\sqrt{3}+18}{3}\left[\text{dm}\right]$  

Pole:

$P=\frac{1}{2}\cdot \left|BC\right|\cdot \left|AC\right|=\frac{1}{2}\cdot 2\sqrt{3}\cdot 6=6\sqrt{3}\left[{\text{dm}}^2\right]$