How do you find a side of a scalene triangle with internal angles and area?

Let $A$ be the area of a triangle, $x,y,z$ be the three sides and $X,Y,Z$ be their opposite angles. Then 
\[ (1)    A=\frac{1}{2}xy \sin Z  \Leftrightarrow   xy=\frac{2A}{\sin Z}\]
\[(2)     A=\frac{1}{2}yz \sin X  \Leftrightarrow yz=\frac{2A}{\sin X}\]
and
\[(3)     A=\frac{1}{2}xz \sin Y \Leftrightarrow xz=\frac{2A}{\sin Y}.\]
Myltiplying these three equations together we get
\[A^3=(\frac{1}{2})^3 (xyz)^2 \sin X \sin Y \sin Z.\]
Hence 
\[xyz= \sqrt{\frac{(2A)^3}{\sin X \sin Y \sin Z}}=\frac{2A\sqrt{2A}}{\sqrt{\sin X \sin Y \sin Z}}\]
\[=\frac{2A}{\sin Z}\frac{\sin Z\sqrt{2A}}{\sqrt{\sin X \sin Y \sin Z}}=\frac{2A}{\sin Z}\sqrt{\frac{2A \sin Z}{\sin X \sin Y}}\]
\[=xy \sqrt{\frac{2A \sin Z}{\sin X \sin Y}}.\]
So
\[ (4)  z=\sqrt{\frac{2A \sin Z}{\sin X \sin Y}}.\]
Similarly we get
\[(5)  y=\sqrt{\frac{2A \sin Y}{\sin X \sin Z}}.\]
and
\[(6)   x=\sqrt{\frac{2A \sin X}{\sin Y \sin Z}}.\]
Hence you can get the sides of the tirangle by simply using the formulas (4), (5), and (6). Indeed 
\[z=\sqrt{\frac{2\times(250) \sin 70}{\sin 62 \sin 48}}= 26.1033m\]
\[y=\sqrt{\frac{2\times(250) \sin 48}{\sin 62 \sin 70}}= 24.9326m\]
\[x=\sqrt{\frac{2\times(250) \sin 62}{\sin 48 \sin 70}}= 25.9132 m.\]