Finding length of a side in a right triangle with these conditions

Question

$\triangle ABC$ is a right triangle ($\measuredangle A = 90^{\circ} $). There is two point $D , E$ on $BC$ such that $BD = DE = EC$. We also know that $AD = \sin x,\,$ and $AE =\cos x\,$ such that ($0^{\circ}

I must note that the length of BD will be a real constant number and it is not something parametric like $BD = a + b -c$.

Answer 1

We may suppose that $$A(0,0),\quad B(0,b),\quad C(c,0)$$ where $b,c\gt 0$.

Then, we get $$D\left(\frac c3,\frac{2b}{3}\right),\quad E\left(\frac{2c}{3},\frac b3\right)$$ from which we have $$AD^2=\frac{c^2+4b^2}{9},\quad AE^2=\frac{4c^2+b^2}{9}$$ Since $\sin^2x+\cos^2x=1$, we have $$1=AD^2+AE^2=\frac{5b^2+5c^2}{9}\implies b^2+c^2=\frac 95$$ Therefore, $$BD=\sqrt{\frac{c^2+b^2}{9}}=\sqrt{\frac 19\times\frac 95}=\color{red}{\frac{1}{\sqrt 5}}$$

Answer 2

A bit of trigonometry:

Let $BC= l$, $\measuredangle BCA =\gamma$.

Cosine law:

1) Triangle $ADC$:

$ \sin ^2 (x) = (AC)^2 + \left(\frac{2}{3} l\right)^2 - 2AC (\frac{2}{3})l \cos (\gamma) $

2) Triangle $AEC$:

$\cos ^2 (x) = AC^2 + (\frac{1}{3}l)^2 - 2AC(\frac{1}{3} l) \cos (\gamma)$.

An equation for $\gamma$:

3) $ \cos (\gamma) = \frac{AC}{l} $ , since triangle $CAB$ is a right triangle.

Replace$ \cos (\gamma) $ in 1) and 2):

1)$ \sin ^2 (x) = AC^2 + (\frac{2}{3} l)^2 - 2 AC (\frac{2}{3} l) \frac{AC}{l} =$

$= AC^2 - (\frac{4}{3}) AC^2 + (\frac{4}{9})l^2 $.

2)$ \cos ^2 (x) = AC^2 + (\frac{1}{3}l)^2 - 2AC (\frac{1}{3} l)\frac{AC}{l} =$

$= AC^2 - (\frac{2}{3})AC^2 + (\frac{1}{9})l^2 $

Adding 1) and 2):

$1 = (\frac{4}{9}) l^2 + (\frac{1}{9}) l^2$

$1 = (\frac{5}{9}) l^2$, or

$l= (\frac{3}{\sqrt 5})$, and finally

$BD = (\frac{1}{3}) l= \frac{1}{\sqrt5}$.