Answer
The length of the sides of triangle are $a=11.6$ and $b=23.9$
Work Step by Step
Let the lower left point and lower right point of given triangle be $D\text{ and }E$, respectively.
Now by applying law of cosines in triangle $ADE$, we get
$\begin{align}
& \cos A=\frac{{{17}^{2}}+{{26}^{2}}-{{37.5}^{2}}}{2\cdot 17\cdot 26} \\
& \cos A=\frac{289+676-1406.25}{884} \\
& A={{\cos }^{-1}}\left( -.499 \right) \\
& A\simeq 120{}^\circ
\end{align}$
Similarly, we will apply the law of cosines in triangle $ABE$
$\begin{align}
& \cos A=\frac{{{26}^{2}}+{{21.5}^{2}}-{{13.5}^{2}}}{2\cdot 26\cdot 21\cdot 5} \\
& \cos A=\frac{676+462.25-182.25}{1118} \\
& A={{\cos }^{-1}}.855 \\
& A\simeq 31{}^\circ
\end{align}$
After that we will apply the sine law
$\begin{align}
& \frac{13.5}{\sin 31{}^\circ }=\frac{21.5}{\sin E} \\
& E=56{}^\circ \\
\end{align}$
And
$\begin{align}
& B=180{}^\circ -56{}^\circ -31{}^\circ \\
& B=93{}^\circ \\
\end{align}$
Now, consider the triangle $ACB$, to find out the angle $\angle CAB$
$\begin{align}
& \angle CAB=180-\angle BAE-\angle EAD \\
& \angle CAB=180{}^\circ -120{}^\circ -31{}^\circ \\
& \angle CAB=29{}^\circ \\
\end{align}$
Consider the triangle $ACB$, to find out the angle $\angle CBA$
$\begin{align}
& \angle CBA=180-\angle EBA \\
& \angle CAB=180{}^\circ -93{}^\circ \\
& \angle CAB=87{}^\circ \\
\end{align}$
And
$\begin{align}
& C=180{}^\circ -87{}^\circ -29{}^\circ \\
& C=64{}^\circ \\
\end{align}$
Now by using sine law we get,
$\begin{align}
& \frac{a}{\sin 29{}^\circ }=\frac{21.5}{\sin 64{}^\circ } \\
& a\simeq 11.6
\end{align}$
Again, by applying sine law we will find out the side of the triangle.
$\begin{align}
& \frac{b}{\sin 87{}^\circ }=\frac{21.5}{\sin 64{}^\circ } \\
& b\simeq 23.9
\end{align}$
