Respuesta :

Since the tangent of B is 4/3, then:

[tex]\begin{gathered} \frac{AC}{BA}=\frac{4}{3} \\ \Rightarrow AC=\frac{4}{3}BA \end{gathered}[/tex]

On the other hand, since ABC is a right triangle with hypotenuse BC, then:

[tex](BA)^2+(AC)^2=(BC)^2[/tex]

Substitute AC=3/4 BA and BC=15 to find BA:

[tex]\begin{gathered} (BA)^2+(\frac{4}{3}BA)^2=15^2 \\ \Rightarrow BA^2+\frac{16}{9}BA^2=15^2 \\ \Rightarrow(1+\frac{16}{9})BA^2=15^2 \\ \Rightarrow\frac{25}{9}BA^2=15^2 \\ \Rightarrow BA^2=\frac{9}{25}\cdot15^2 \\ \Rightarrow BA=\sqrt[]{(\frac{9}{25}\cdot15^2)} \\ \Rightarrow BA=\frac{3}{5}\cdot15 \\ \Rightarrow BA=9 \end{gathered}[/tex]

Substitute BA=12 into the expression for AC to find its value:

[tex]AC=\frac{4}{3}BA=\frac{4}{3}(9)=12[/tex]

On the other hand, we know that:

[tex]BD+DA=BA[/tex]

Substitute BA=12 and DA=3 to find BD:

[tex]\begin{gathered} BD+3=9 \\ \Rightarrow BD=9-3 \\ \Rightarrow BD=6 \end{gathered}[/tex]

Finally, since BDE and BAC are similar triangles, we know that:

[tex]\begin{gathered} \frac{DE}{BD}=\frac{AC}{BA} \\ \Rightarrow DE=\frac{AC}{BA}\times BD \end{gathered}[/tex]

Substitute AC=12, BA=9 and BD=6 to find the length DE:

[tex]\begin{gathered} DE=\frac{12}{9}\times6 \\ \Rightarrow DE=8 \end{gathered}[/tex]

Therefore, the length of DE is:

[tex]8[/tex]

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