Answer
$230.9$
Work Step by Step
Let $x=x_{1}+x_{2}$
where $x_{1}$ is the leg adjacent to $60^{o}$in the right triangle on the left,
and $x_{2} $is the leg adjacent to $30^{o}$ in the right triangle on the right.
$\displaystyle \tan 60^{o}=\frac{100}{x_{1}}\ \Rightarrow\ x_{1}=\frac{100}{\tan 60^{o}}$
$\displaystyle \tan 30^{o}=\frac{100}{x_{2}}\ \Rightarrow\ x_{2}=\frac{100}{\tan 30^{o}}$
$x=\displaystyle \frac{100}{\tan 60^{o}}+\frac{100}{\tan 30^{o}}\approx 230.9$
