Question

Calcule a medida do lado BC do triângulo ABC

Answer 1

Solução:

Por meio da Lei dos senos, temos que

$\dfrac{BC}{sen45\°}=\dfrac{AB}{sen30\°}$

Onde $AB=3\sqrt{2}$, $sen45\°=\dfrac{\sqrt{2}}{2}$ e $sen30\°=\dfrac{1}{2}$, logo

$\dfrac{BC}{\frac{\sqrt{2}}{2}}=\dfrac{3\sqrt{2}}{\frac{1}{2}}\\\\\\\dfrac{2}{\sqrt{2}}\cdot{BC}=\dfrac{2}{1}\cdot3\sqrt{2}\\\\\\\dfrac{\not{2}}{\sqrt{2}}\cdot BC=\not{2}\cdot 3\sqrt{2}\\\\\\\dfrac{BC}{\sqrt{2}}=3\sqrt{2}\\\\\\BC=3\sqrt{2}\sqrt{2}\\\\\\BC=3\sqrt{2\cdot2}\\\\\\BC=3\sqrt{4}\\\\\\BC=3\cdot2\\\\\\\boxed{\boxed{BC=6}}$