Hello, Jamie!

[quote] Code:

A o
* *
x *50°*
* *
* *
D o *
* * *
* * 10°*
* * *
* 30° * *
C o * * * * * * * * * o B
100

In right triangle $\displaystyle DCB\!:\;\;\tan30^o \:=\:\frac{CD}{100}\quad\Rightarrow\quad CD \:=\:100\tan30^o\;\;{\color{blue}[1]}$

In right triangle $\displaystyle ACB\!:\;\;\tan40^o \:=\:\frac{AC}{100}\quad\Rightarrow\quad AC \:=\:100\tan40^o\;\;{\color{blue}[2]}$

But $\displaystyle AC \:=\:x + CD\quad\Rightarrow\quad x \:=\:AC - CD\;\;{\color{blue}[3]}$

Substitute [1] and [2] into [3]: . $\displaystyle \boxed{x \;=\;100\tan40^o - 100\tan30^o}$