To start with, the hypotenuse for the triangle with $\displaystyle \begin{align*} x \end{align*}$ in it is the same as the side opposite angle $\displaystyle \begin{align*} \theta \end{align*}$ in the upper triangle. Call this side $\displaystyle \begin{align*} y \end{align*}$ and we have
$\displaystyle \begin{align*} \tan{\theta} &= \frac{y}{10} \\ y &= 10\tan{\theta} \end{align*}$
As for the third side of the triangle with $\displaystyle \begin{align*} x \end{align*}$ in it, call this side $\displaystyle \begin{align*} z \end{align*}$. If you complete the rectangle, you'll have another right angle triangle with that side opposite $\displaystyle \begin{align*} \theta \end{align*}$ and with a hypotenuse of $\displaystyle \begin{align*} 10 \end{align*}$. That means
$\displaystyle \begin{align*} \sin{\theta} &= \frac{z}{10} \\ z &= 10\sin{\theta} \end{align*}$
So finally, in the triangle with $\displaystyle \begin{align*} x \end{align*}$ in it, we have by Pythagoras
$\displaystyle \begin{align*} x^2 + \left(10\sin{\theta}\right)^2 &= \left(10\tan{\theta}\right)^2 \\ x^2 + 100\sin^2{\theta} &= 100\tan^2{\theta} \\ x^2 &= 100\tan^2{\theta} - 100\sin^2{\theta} \\ x^2 &= 100\left(\tan^2{\theta} - \sin^2{\theta}\right) \\ x^2 &= 100\left(\frac{\sin^2{\theta}}{\cos^2{\theta}} - \sin^2{\theta}\right) \\ x^2 &= 100\left( \frac{\sin^2{\theta}}{\cos^2{\theta}} - \frac{\sin^2{\theta}\cos^2{\theta}}{\cos^2{\theta} } \right) \\ x^2 &= 100\left( \frac{\sin^2{\theta} - \sin^2{\theta}\cos^2{\theta}}{\cos^2{\theta}} \right) \\ x^2 &= 100\left[ \frac{\sin^2{\theta}\left( 1 - \cos^2{\theta} \right)}{\cos^2{\theta}} \right] \\ x^2 &= 100\left( \frac{\sin^2{\theta}\sin^2{\theta}}{\cos^2{\theta} } \right) \\ x^2 &= 100\left( \frac{\sin^4{\theta}}{\cos^2{\theta}} \right) \\ x &= 10\left( \frac{\sin^2{\theta}}{\cos{\theta}} \right) \\ x &= 10\sin{\theta}\tan{\theta} \end{align*}$
I think that's the simplest you can get it