Matemática, perguntado por jv321239, 5 meses atrás

Calcule o seno, cosseno e tangente do ângulo alfa

Anexos:

Soluções para a tarefa

Respondido por CyberKirito
2

\large\boxed{\begin{array}{l}\rm x\longrightarrow hipotenusa\,do\,tri\hat angulo\\\rm x^2=(\sqrt{3})^2+(\sqrt{5})^2\\\rm x^2=3+5\\\rm x^2=8\\\rm x=\sqrt{8}\\\rm x=\sqrt{4\cdot2}\\\rm x=2\sqrt{2}\\\rm sen(\alpha)=\dfrac{\sqrt{3}}{x}=\dfrac{\sqrt{3}}{2\sqrt{2}}\\\\\rm sen(\alpha)=\dfrac{\sqrt{3}\cdot\sqrt{2}}{2\cdot\sqrt{2^2}}=\dfrac{\sqrt{6}}{4}\\\\\rm cos(\alpha)=\dfrac{\sqrt{5}}{2\sqrt{2}}\\\\\rm cos(\alpha)=\dfrac{\sqrt{5}\cdot\sqrt{2}}{5\cdot\sqrt{2^2}}\end{array}}

\large\boxed{\begin{array}{l}\rm cos(\alpha)=\dfrac{\sqrt{10}}{2\sqrt{5^2}}=\dfrac{\sqrt{10}}{10}\\\\\rm tg(\alpha)=\dfrac{\sqrt{3}}{\sqrt{5}}\\\\\rm tg(\alpha)=\dfrac{\sqrt{3}\cdot\sqrt{5}}{\sqrt{5^2}}=\dfrac{\sqrt{15}}{5}\end{array}}

Perguntas interessantes