Question

Resolva as Razões Trigonométricas
Em cada triângulo, determine os valores x , sen a, cos a , tg a.

Answer 1

$\Large\boxed{\begin{array}{l}\rm a)\\\sf x^2+3^2=5^2\\\sf x^2+9=25\\\sf x^2=25-9\\\sf x^2=16\\\sf x=\sqrt{16}\\\sf x=4\\\sf sen(\alpha)=\dfrac{4}{5}\\\\\sf cos(\alpha)=\dfrac{3}{5}\\\\\sf tg(\alpha)=\dfrac{4}{3}\end{array}}$

$\Large\boxed{\begin{array}{l}\rm b)\\\sf x^2+12^2=13^2\\\sf x^2+144=169\\\sf x^2=169-144\\\sf x^2=25\\\sf x=\sqrt{25}\\\sf x=5\\\sf sen(\alpha)=\dfrac{12}{13}\\\\\sf cos(\alpha)=\dfrac{5}{13}\\\\\sf tg(\alpha)=\dfrac{12}{5}\end{array}}$

$\Large\boxed{\begin{array}{l}\rm c)\\\sf x^2=2\cdot7^2\\\sf x=\sqrt{2\cdot7^2}\\\sf x=7\sqrt{2}\\\sf sen(\alpha)=cos(\alpha)=\dfrac{\diagdown\!\!7}{\diagdown\!\!\!\!7\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\\\\sf tg(\alpha)=\dfrac{7}{7}=1\end{array}}$

$\Large\boxed{\begin{array}{l}\rm d)\\\sf x^2= 36^2+(12\sqrt{3})^2\\\sf x^2=1296+432\\\sf x^2=1728\\\sf x=\sqrt{1728}\\\sf x=24\sqrt{3}\\\sf sen(\alpha)=\dfrac{\diagdown\!\!\!\!\!\!36^3}{\diagdown\!\!\!\!\!\!24_{2}\sqrt{3}}=\dfrac{\diagup\!\!\!3\sqrt{3}}{2\cdot\diagup\!\!\!\!3}=\dfrac{\sqrt{3}}{2}\\\\\sf cos(\alpha)=\dfrac{\diagdown\!\!\!\!\!\!12~\diagup\!\!\!\!\!\sqrt{3}}{\diagdown\!\!\!\!\!24~\diagup\!\!\!\!\!\sqrt{3}}=\dfrac{1}{2}\\\\\sf tg(\alpha)=\dfrac{36}{12\sqrt{3}}=\dfrac{36\sqrt{3}}{12\cdot3}=\sqrt{3}\end{array}}$