Na figura abaixo, sabe-se que $|\overrightarrow{A B}|=|\overrightarrow{B C}|=26$ e $|\overrightarrow{A C}|=48$. No triângulo $A B C$ tem-se que $\overrightarrow{D B} \perp \overrightarrow{A C}$ Fonte: elaborado pelo autor (2023). Considerando a situação apresentada, o valor de $|\overrightarrow{D B}|$ em unidades de comprimento, é igual a Alternativas

Step 1 ：First, we draw a diagram. We extend $\overrightarrow{CB}$ and $\overrightarrow{DA}$ to meet at $E$. This gives us a couple right triangles in $CED$ and $BEA$.

Step 2 ：We see that $\angle E = 60^\circ$. Hence, $\triangle BEA$ and $\triangle DEC$ are 30-60-90 triangles.

Step 3 ：Using the side ratios of 30-60-90 triangles, we have $BE=2BA=2*26=52$. This tells us that $CE=BC+BE=26+52=78$. Also, $EA=26\sqrt{3}$.

Step 4 ：Because $\triangle DEC\sim\triangle BEA$, we have $\frac{78}{26\sqrt{3}}=\frac{CD}{26}$.

Step 5 ：Solving the equation, we have \begin{align*} \frac{CD}{26}&=\frac{78}{26\sqrt{3}}\\ CD&=26\cdot\frac{78}{26\sqrt{3}}\\ CD&=\boxed{\frac{78}{\sqrt{3}}} \end{align*}