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Image figure available here

In the figure above, tan B = $\frac{3}{4}$. If BC = 15 and DA = 4, what is the length of DE?

Source: Khanacademy.org Please help me solve this question! I don't know where to start and I can't find anything. Please help me!

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    $\begingroup$ What have you tried? Which sides are you able to find the length of? $\endgroup$ – Arthur Aug 28 at 10:59
  • $\begingroup$ We can help if you tell us where you are stuck exactly $\endgroup$ – Klangen Aug 28 at 11:07
  • $\begingroup$ Sorry! I don't have any idea of where to start, and I'm completely stuck! $\endgroup$ – Felicia Natalia Rivera Aug 30 at 0:55
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Let $x$ the lenght of $AC$ and $y$ the lenght of $BD$, so I have: $$\left\{\begin{matrix} x^2+(y+4)^2=15^2 \\ \frac{x}{y+4}=\frac{3}{4} \end{matrix}\right.$$

That is the same as:

$$\left\{\begin{matrix} y^2+8y-128=0 \\x=\frac{3}{4}\cdot(y+4) \end{matrix}\right.$$

This has solutions: $x=9$ and $y=8$. The two right triangle are similar, so: $\frac{BD}{AB}=\frac{DE}{AC}$. From this, I obtain: $\frac8{12}=\frac{DE}{9}$ and $DE=6$.

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AC=3x 
AB=4x
(3x)²+(4x)²=225 
x=3 
BD=8
AB/AC=BD/DE
12/9=8/DE
DE=6
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