<s>[INST] Can one always map a given triangle into a triangle with chosen angles by means of a parallel projection? This is something that seems to be true from experience by playing with shadows from the sun:
If one cuts a paper triangle, he can turn it in a way to make its shadow be a triangle of any given angles (of course, not exceeding internal sum of 180°), for example: If one draws a right triangle with internal angles 90°, 60° and 30°,  one can turn it in a way that the shadow triangle can be of angles 60°, 60°, 60° (equilateral) or 120°, 30° and 30° (isosceles), or any other triangle possibilities.
Is it true? How can we prove it? (if possible, showing not only that it's possible, but how to obtain the specific mapping that does it)
 [/INST] We'll take the special case of projection along the $z$-axis to a target triangle $\triangle A^\prime B^\prime C^\prime$ inscribed in the unit circle. Without loss of generality, we can take the triangle $\triangle ABC$ making the projection to be
$$A = (1,0,0) \qquad B = (\cos2C^\prime,\sin2C^\prime, p) \qquad C = (\cos2B^\prime, -\sin2B^\prime, q)$$
for some $p$ and $q$. (The counterpart $z$ coordinate of $A$ vanishes with an appropriate vertical translation.) In accordance with the Law of Sines, we have
$$\frac{|\overline{BC}|}{\sin A} = \frac{|\overline{CA}|}{\sin B} = \frac{|\overline{AB}|}{\sin C}$$
which gives a non-linear system in $p$ and $q$:
$$\begin{align}
p^2 \sin^2 A - ( p - q )^2 \sin^2 C + 4 \sin^2 A\sin^2C^\prime - 4 \sin^2A^\prime \sin^2 C &= 0 \\
q^2\sin^2 A - ( p - q )^2 \sin^2 B + 4 \sin^2 A \sin^2 B^\prime - 4 \sin^2 A^\prime \sin^2B &= 0
\end{align}$$
Eliminating $q$ from the system is straightforward enough, yielding a quadratic in $p^2$:
$$\begin{align}
0 \quad=\quad &p^4 \sin^2 A \sin^2 B \\
- &p^2  \left(\begin{array}{c}\left(\;\sin\alpha^{+} \sin\beta^{-} - \sin\alpha^{-} \sin\beta^{+} \;\right)^2 \\
- 4 \sin^2A \sin^2B \sin^2C^\prime + 4 \sin^2A^\prime \sin^2B^\prime \sin^2 C\;) \end{array}\right) \\
- &4 \sin^2C^\prime \left(\;\sin\alpha^{+} \sin\beta^{-} - \sin\alpha^{-} \sin\beta^{+} \;\right)^2
\end{align}$$
where
$$\alpha^{\pm} = A \pm A^\prime \qquad \beta^{\pm} = B \pm B^\prime \qquad \gamma^{\pm} = C\pm C^\prime$$
Because this quadratic has a non-negative leading term, and a non-positive constant term, Descartes' Rule of Signs guarantees that it has exactly one positive real root, so we can unambiguously write
$$p^2 = \frac{\left(\begin{array}{c}\left(\;\sin\alpha^{+} \sin\beta^{-} - \sin\alpha^{-} \sin\beta^{+} \;\right)^2 \\
- 4 \sin^2A \sin^2B \sin^2C^\prime + 4 \sin^2A^\prime \sin^2B^\prime \sin^2 C\;) \end{array}\right) + \sqrt{\Delta}}{2\sin^2 A \sin^2 B}$$
where 
$$\begin{align}
\Delta = \left(\;
\begin{array}{c}
    \sin^2\alpha^{-} + \sin^2\beta^{-} + \sin^2\gamma^{-} \\
+ 2 \cos\alpha^{+} \sin\beta^{-} \sin\gamma^{-} \\
+ 2 \sin\alpha^{-} \cos\beta^{+} \sin\gamma^{-} \\
+ 2 \sin\alpha^{-} \sin\beta^{-} \cos\gamma^{+} 
\end{array}
\;\right) \;\cdot \left(\;
\begin{array}{c}
    \sin^2\alpha^{+} + \sin^2\beta^{+} + \sin^2\gamma^{+} \\
+ 2 \cos\alpha^{-} \sin\beta^{+} \sin\gamma^{+} \\
+ 2 \sin\alpha^{+} \cos\beta^{-} \sin\gamma^{+} \\
+ 2 \sin\alpha^{+} \sin\beta^{+} \cos\gamma^{-} 
\end{array}
\;\right)
\end{align}$$
The corresponding equation for $q^2$ arises by exchanging the roles $B\leftrightarrow C$, $B^\prime \leftrightarrow C^\prime$, $\beta^{\pm}\leftrightarrow\gamma^{\pm}$. (Note that $\Delta$ remains unchanged.)
Determination of the signs of $p$ and $q$ themselves is left as an exercise to the reader. 
</s>