Question

Alguém poderia me ajudar com esse calculo

Answer 1

Olá, Nasantos.

$\cos30\º=\frac{AC}{18}\Rightarrow \frac{\sqrt3}{2}=\frac{AC}{18}\Rightarrow AC=\frac{18}{2}\sqrt3=9\sqrt3\,cm\\\\ \tan60\º=\frac{30}{BC}\Rightarrow\sqrt3=\frac{30}{BC}\Rightarrow BC=\frac{30}{\sqrt3}\cdot\frac{\sqrt3}{\sqrt3}=10\sqrt3\,cm\\\\ \therefore AB=AC+BC=9\sqrt3+10\sqrt3=\boxed{19\sqrt3\,cm}$