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)

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. 
