Zhadnyy,
- 20200522_150046.jpg (14.21 KiB) Exibido 758 vezes
[tex3]\senθ=\frac{b}{\ell}\implies\cosθ=\sqrt{1-\frac{b^2}{\ell^2}}=\frac{1}{\ell}\sqrt{\ell^2-b^2}\\
\sen(60°-θ)=\frac{a}{\ell}\to\frac{\sqrt3}{2}\cdot\frac{1}{\ell}\sqrt{\ell^2-b^2}-\frac{b}{\ell}\cdot\frac12=\frac{a}{\ell}[/tex3]
Fazendo as continhas, vem que:
[tex3]\ell=2\sqrt{(a^2+ab+b^2)/3}[/tex3]
Dias de luta, dias de glória.