Question

Determine Sen Â, no triângulo a seguir:
(IMAGEM EM ANEXO!!!)

Answer 1

Pela lei dos senos:

$\dfrac{a}{\texttt{sen} \ A}=\dfrac{b}{\texttt{sen} \ B}=\dfrac{c}{\texttt{sen} \ C}$

$\dfrac{AC}{\texttt{sen} \ B}=\dfrac{BC}{\texttt{sen} \ A}$

$\dfrac{10\sqrt{2}}{\texttt{sen} \ 45^{\circ}}=\dfrac{5}{\texttt{sen} \ A}$

$\dfrac{10\sqrt{2}}{\frac{\sqrt{2}}{2}}=\dfrac{5}{\texttt{sen} \ A}$

$10\sqrt{2}\cdot\dfrac{2}{\sqrt{2}}}=\dfrac{5}{\texttt{sen} \ A}$

$20=\dfrac{5}{\texttt{sen} \ A}$

$\texttt{sen} \ A}=\dfrac{5}{20} \iff \boxed{\texttt{sen} \ A=\dfrac{1}{4}}$