Question

No triângulo ABC, abaixo, tem-se que tg a = 2. Qual valor da medida do lado BC?

Answer 1

$\sf tg(\alpha)=\dfrac{BA}{AC}\\\sf 2=\dfrac{BA} {x}\\\sf BA=2x\\\sf (2x+1)^2=x^2+(2x)^2\\\sf \diagdown\!\!\!\!\!4x^2+4x+1=x^2+\diagdown\!\!\!\!\!4x^2\\\sf x^2-4x-1=0\\\sf\Delta=b^2-4ac\\\sf\Delta=(-4)^2-4\cdot1\cdot(-1)\\\sf\Delta=16+4\\\sf\Delta=20\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\sf x=\dfrac{-(-4)\pm\sqrt{20}}{2\cdot1}\\\sf x=\dfrac{4\pm2\sqrt{5}}{2}\\\sf x=\dfrac{\diagdown\!\!\!\!2\cdot(2\pm\sqrt{5})}{\diagdown\!\!\!\!2}\\\sf x=2\pm\sqrt{5}\begin{cases}\tt x_1=2+\sqrt{5}\\\tt x_2=2-\sqrt{5}\end{cases}\\\sf como~2-\sqrt{5}<0\\\sf adotaremos~somente~a~parte~positiva.\\\boxed{\rm x=2+\sqrt{5}}\\\sf BC=2\cdot x\\\large\boxed{\boxed{\boxed{\boxed{\sf BC=2\cdot(2+\sqrt{5})}}}}\blue{\checkmark}$