Question

Determine o valor de M,sendo m um número real, tal que senB=M/6 e Cos B+ √(4m/3)

Answer 1

$\text{sen}^2~\text{B}+\text{cos}^2~\text{B}=1$

$\left(\dfrac{m}{6}\right)^2+\left(\sqrt{\dfrac{4m}{3}}\right)^2=1$

$\dfrac{m^2}{36}+\dfrac{4m}{3}=1$

$m^2+48m-36=0$

$\Delta=48^2-4\cdot1\cdot(-36)=2304+144=2448$

$m=\dfrac{-48+\sqrt{2448}}{2\cdot1}=\dfrac{-48+12\sqrt{17}}{2}=-24+6\sqrt{17}$

$m=6\cdot\left(\sqrt{17}-4\right)$