Fix one side, say segment $AB$, then draw (one arc of) the circle $c$ which consists of those points from where $AB$ is seen in the given angle. So that the centre of the parallelogramma has to be on $c$. Then enlarge $c$ to its double from center $A$, getting arc of circle $c'$ which must contain a $3$rd vertex of the parallelogramma, so finally just draw the circle centered at $B$ with the length of the other side.
Let the parallelepiped with a vertex $O$ at the origin have vertices $A$, $B$, $C$. Write $D = B+C$; $E= C+A$; $F=A+B$; and $G=A+B+C$. Further, assign these names to the lengths of edges and diagonals:
$$a := |OA|\quad b := |OB| \quad c := |OC| \quad d := |AD| \quad e := |BE| \quad f := |CF| \quad g := |OG|$$
Note that a parallelepiped (and it properties) is completely determined by six quantities: the lengths of three edges meeting at a vertex, and the measures of the three angles between those edges. The total number of edge-lengths and diagonal-lengths is seven, so there's a dependency among these values, namely
$$4 \;(\; a^2 + b^2 + c^2\;) = d^2 + e^2 + f^2 + g^2$$
Although one diagonal (say, $g$) is unnecessary, it helps to simplify the volume formula a bit:
$$\begin{align}
32 V^2 &= 32 a^2 b^2 c^2 + d^2 e^2 f^2 + g^2 ( d^2 e^2 + e^2 f^2 + f^2 d^2 )\\
&- 2 g^2 ( a^2 d^2 + b^2 e^2 + c^2 f^2 ) - 2 ( a^2 e^2 f^2 + d^2 b^2 f^2 + d^2 e^2 c^2 )
\end{align}$$
In the case of a rectangular parallelepiped, for which $d^2=e^2=f^2=g^2=a^2+b^2+c^2$, the formula reduces to $V = a b c$, as expected.
I derived the formulas using coordinates
$$A = (a_x,0,0) \qquad B = (b_x, b_y, 0) \qquad C = (c_x, c_y, c_z)$$
Expressing the various distances in terms of these
$$a^2 = A\cdot A = a_x^2 \qquad b^2 = B\cdot B = b_x^2 + b_y^2 \qquad c^2 = C\cdot C = c_x^2 + c_y^2 + c_z^2$$
$$d^2 = \overrightarrow{AD}\cdot \overrightarrow{AD}=\dots \quad e^2 = \overrightarrow{BE}\cdot \overrightarrow{BE} =\dots \quad f^2 = \overrightarrow{CF}\cdot \overrightarrow{CF}=\dots$$
along with
$$V = a_x b_y c_z$$
I systematically eliminated $a_x$, $b_x$, $b_y$, $c_x$, $c_y$, $c_z$ from the system with the help of Mathematica's Resultant[] function. (Appropriate applications of the Law of Cosines should work just as well, but the Resultant process lets me crank through the equations without having to think too hard. :)
An $n$-dimensional parallelotope is determined by $\frac{1}{2}n(n+1)$ values. (These are the triangular numbers. You can see their relevance by noting the coordinatization done above considers vectors using $1$, $2$, $3$, ..., $n$ non-zero coordinates.) The figure has n characteristic edge-lengths and $2^{n-1}$ diagonals, so expressing volume in terms of these is certainly possible. With an increasing number of "extra" lengths, the resulting formula can take a variety of forms.
Best Answer
You can use the law of cosines. You write the following system: \begin{equation} \begin{cases} 80^2=x^2+y^2-2xy\cos120°\\60^2=x^2+y^2-2xy\cos60° \end{cases} \end{equation} where $x$=$AX$ and $y$=$BX$.
Subtracting the equations: \begin{equation} 80^2-60^2=x^2+y^2-2xy\cos120°-(x^2+y^2-2xy\cos60°) \end{equation} we have:
$2800=-2xy(\cos120°-\cos60°)$
$2800=-2xy(-\frac{1}{2}-\frac{1}{2})$
$2800=2xy$
so $xy=1400$
Now you can just calculate the area of the two triangles ABX and BCX using:
\begin{equation} Area=\frac{1}{2}xy\sin\alpha, \end{equation}
where $\alpha$ is the angle between $x$ and $y$ (120° in ABX and 60° in BCX). Then it is easy.