Questão 45 - 1ª Fase - Einstein 2026

Vê-se que no triângulo retângulo BCD, tem-se:

$\frac{3}{4}=\mathrm{tg}\left(\alpha \right)=\frac{\mathrm{BD}}{\mathrm{BC}}=\frac{\mathrm{BD}}{8}\Rightarrow \mathrm{BD}=6$

Utilizando o teorema de Pitágoras no triângulo retângulo BCD, tem-se: ${\mathrm{CD}}^2=6^2+8^2=100$; assim, CD = 10.

Utilizando o teorema de Pitágoras no triângulo retângulo CDE, tem-se: ${\left(5\sqrt{5}\right)}^2={10}^2+{\mathrm{DE}}^2$; assim, DE = 5.

Nota-se que $\mathrm{tg}\left(\mathrm{\beta }+\mathrm{\delta }\right)=\frac{5}{10}=\frac{1}{2}$.

Portanto:

$\mathrm{tg}\left(\left(\mathrm{\alpha }+\mathrm{\beta }\right)-\left(\mathrm{\beta }+\mathrm{\delta }\right)\right)=\frac{\mathrm{tg}\left(\mathrm{\alpha }+\mathrm{\beta }\right)-\mathrm{tg}\left(\mathrm{\beta }+\mathrm{\delta }\right)}{1+\mathrm{tg}\left(\mathrm{\alpha }+\mathrm{\beta }\right)·\mathrm{tg}\left(\mathrm{\beta }+\mathrm{\delta }\right)}=\frac{1-{\displaystyle \frac{1}{2}}}{1+1·{\displaystyle \frac{1}{2}}}=\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{3}{2}}}=\frac{1}{3}$

Ou seja, $\mathrm{tg}\left(\mathrm{\alpha }-\mathrm{\delta }\right)=\frac{1}{3}$.

Logo:

$\frac{1}{3}=\mathrm{tg}\left(\mathrm{\alpha }-\mathrm{\delta }\right)=\frac{\mathrm{tg}\left(\mathrm{\alpha }\right)-\mathrm{tg}\left(\mathrm{\delta }\right)}{1+\mathrm{tg}\left(\mathrm{\alpha }\right)·\mathrm{tg}\left(\mathrm{\delta }\right)}=\frac{{\displaystyle \frac{3}{4}}-\mathrm{tg}\left(\mathrm{\delta }\right)}{1+{\displaystyle \frac{3}{4}}·\mathrm{tg}\left(\mathrm{\delta }\right)}=\frac{3-4\mathrm{tg}\left(\mathrm{\delta }\right)}{4+3\mathrm{tg}\left(\mathrm{\delta }\right)}$

Resolvendo a equação $\frac{3-4\mathrm{tg}\left(\mathrm{\delta }\right)}{4+3\mathrm{tg}\left(\mathrm{\delta }\right)}=\frac{1}{3}$, conclui-se que:

$9-12\mathrm{tg}\left(\mathrm{\delta }\right)=4+3\mathrm{tg}\left(\mathrm{\delta }\right)\Rightarrow 15\mathrm{tg}\left(\mathrm{\delta }\right)=5\Rightarrow \boxed{\mathrm{tg}\left(\mathrm{\delta }\right)=\frac{1}{3}}$