Respostas

2015-04-08T19:05:02-03:00
Utilizando a fórmula da tangente:
\tan(60) = \dfrac{cateto\_oposto}{cateto\_adjacente}\\\\
\tan(60)=\dfrac{y_1-y_0}{x_1-x_0}\\\\
\tan(60)=\dfrac{2-k}{0-\sqrt{3}}\\\\
\sqrt{3} \times \tan(60) - 2=-k\\
k=\sqrt{3} \times \tan(60) + 2\\
k=\sqrt{3} \times \sqrt{3} + 2\\
k=\sqrt{3}^2 + 2\\
k=3 + 2\\
\boxed{k=5}
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