construct a triangle with a compass and a ruler, given $a, B, t_a$ If $a, b, c$ are the side lengths of a triangle, $A,B,C$ are the opposite internal angles, respectively, and $t_a, t_b, t_c$ are internal bisectors of the angles $A,B,C$, how could I construct a triangle with a compass and straightedge given $a, B, t_a$?

Thank you very much! 
I've tried searching here, which lists multiple ways of constructing a triangle.
 A: No such construction is possible in general, because it wold amount to
a compass-and-straightedge solution of a sextic that is generally
irreducible.  For example, there is no compass-and-straightedge
construction of a triangle with $a=t_a$ and $B = 60^\circ$.
[Added later: a more amusing example is $B = 90^\circ$ with
$t_a/a = \sqrt 2$ or $1/2$, for which the problem is equivalent
to (of all things) construction of a regular heptagon.]
Let $a,b,c$ be the sides of the triangle.  We're given $a$,
$$
\cos B = \frac{a^2+c^2-b^2}{2ac}
$$
(Law of Cosines), and
$$
t_a^2 = \frac{bc((b+c)^2-a^2)}{(b+c)^2}
$$
(see for example
Wikipedia,
noting that $2s$ and $2(s-a)$ are $b+c \pm a$).
This gives simultaneous equations in $b,c$ of degree $2$ and $3$;
taking resultants to eliminate one variable gives a sextic polynomial
in the other.  I claim that this polynomial is generally irreducible.
This is easy to prove computationally by just giving an example.
Suppose that $B$ is a $60$-degree angle and $a=t_a=1$.  Then
$\cos B = 1/2$, so the Law of Cosines gives $b^2 = c^2 - c + 1$,
and we soon find that $c$ is a root of $3c^6-3c^5-c^4+4c^3-3c^2-2c+1$,
which is irreducible (indeed with Galois group $S_6$, the largest possible
given this degree).  Hence $c$ is not constructible, QED.
[Added later: Alternatively, if we take $B = 90^\circ$ then $b^2 = a^2 + c^2$
and we get a cubic in $c^2$, which is simpler than a random sextic in $c$
but still generally unsolvable; an amusing example is $a=1$, $t_a = \sqrt 2$,
for which $c^2 = 2 \cos (\pi/7)$ so constructing such a triangle
would be equivalent with the famously impossible construction of
a regular heptagon!  Likewise $t_a = 1/2$ yields
$c^2 = \cos(3\pi/7) - \cos^2(3\pi/7)$ so again it comes down to
the unconstructible regular heptagon.]
