I'll describe my methods:
1: First of all $∠BAC = 50^\circ$. Given that $AP$ is the angle bisector of $∠A$, we conclude that $∠BAP = 25^\circ$ and $∠CAP=25^\circ$.
2:We know that in a circle the angles formed by a chord on the circumference equal one another (proof in the image here: http://en.wikipedia.org/wiki/Inscribed_angle#Theorem).
Therefore, considering chord $BP$, we have $∠BAP=∠BRP$, or $∠BRP=25^\circ$. Same way, considering chord $PC$, we have $∠CQP = 25^\circ$. With the same methods, we get $∠CQR = \frac{∠B}{2}$ and $∠BRQ = \frac{∠C}{2}$.
3: Doing this with $∠QPR$, we get $∠APQ =\frac{∠C}{2}$ and $∠APR = \frac{∠B}{2}$. So, $$∠QPR = \frac{∠B}{2} + \frac{∠C}{2} = \frac{∠B+∠C}{2} = \frac{180^\circ-∠A}{2} = 90^\circ - \frac{∠A}{2} = 65^\circ$$
Similarly, we have $∠PQR= 90^\circ -\frac{∠C}{2}$ and $∠PRQ = 90^\circ-\frac{∠B}{2}$.
At this point it would be good to realize that we cannot calculate all the angles you asked for, however, as we saw, we can calculate it at least in terms of other angles. We can definitely not calculate $∠C$ and $ ∠B$ from the information, as Nicolas said in the comments. The only restriction which lies on them is that $∠B+∠C = 130^\circ$ [angle-sum property], which can be used to calculate the other when one is given. And if we are given one of them, you could substitute them into the formulas we derived, to obtain the results.
4: Now, $∠QBP = ∠QRP = 90^\circ - \frac{∠B}{2}$, $∠BQP =∠PAB =25^\circ$, $∠BPQ = ∠BCQ = \frac{∠C}{2}$. Remember, the rule I told in the 2nd point. These follow directly from it. Using that rule, you can calculate any other angle you want.
As mentioned in a comment, the path to a polynomial relationship between area and angle bisectors is straightforward (though potentially computationally-expensive) using, say, the method of resultants or Groebner bases to eliminate side-lengths $a$, $b$, $c$ from the system
$$\begin{align}
d^2 &= \frac{bc}{(b+c)^2}((b+c)^2-a^2) \\[4pt]
e^2 &= \frac{ca}{(c+a)^2}((c+a)^2-b^2) \\[4pt]
f^2 &= \frac{ab}{(a+b)^2}((a+b)^2-c^2) \\[4pt]
16 t^2 &= (a+b+c)(-a+b+c)(a-b+c)(a+b-c)
\end{align}$$
where $d$, $e$, $f$ are the lengths of angle bisectors and $t$ is the area of the triangle.
My laptop with Mathematica struggles with the elimination process. There could be ways to optimize, but it turns out that I don't have to work that hard.
The 2005 paper "Area of a Triangle and Angle Bisectors" (PDF link via arXiv.org) by Buturlakin, et al. (2005), discusses how the area of a triangle isn't expressible in terms of the angle bisectors using radicals. Although it doesn't give an explicit polynomial relationship between area and bisectors, it does give relationships involving inradius $r$.
Let us define
$$s_2 = \frac1{d^2}+\frac1{e^2}+\frac1{f^2} \qquad s_3 = \frac1{def} \qquad s_4= \frac1{d^2e^2}+\frac1{e^2f^2}+\frac1{f^2d^2}$$
Then we have
$$4s_2r^2t^2 - 8 s_3 r^3 t^2 = r^4 + t^2 \tag{1}$$
attributed to van Renthe Fink (1843), and
$$\begin{align}
0
&= \phantom{1}64r^{10} s_3^2 (s_2^2 - 4 s_4) \\
&- \phantom{1}64r^9 s_3 (s_2^3 - 10 s_3^2 - 4 s_2 s_4) \\
&+ \phantom{1}16r^8 s_2 (s_2^3 - 50 s_3^2 - 4 s_2 s_4) \\
&+ \phantom{1}32r^7 s_3 (10 s_2^2 - s_4) \\
&- \phantom{19}4r^6 (10 s_2^3 - 61 s_3^2 - 4 s_2 s_4) \\
&-188r^5 s_2 s_3 \\
&+\phantom{1}33r^4 s_2^2 \\
&+\phantom{1}28r^3 s_3 \\
&-\phantom{1}10r^2 s_2 \\
&+\phantom{19}1
\end{align}\tag{2}$$
attributed to H. Wolfe (1937). (The Wolfe polynomial cited in Buturlakin given for $1/(2r)$. I rewrote it for $r$.)
Eliminating $r$ from $(1)$ and $(2)$ is comparatively easy. The result(ant) is ... deep breath ...
$$\begin{align}
0 &= 16777216 t^{20} s_3^{12} (s_2^2 - 4 s_4) \\
&+ 2097152 t^{18} s_3^8 (s_2^6 - 2 s_2^3 s_3^2 - 10 s_3^4 - 6 s_2^4 s_4 +
8 s_2 s_3^2 s_4 + 8 s_2^2 s_4^2) \\[4pt]
&+65536 t^{16} s_3^4 \left(\begin{array}{c}
s_2^{10} + 12 s_2^7 s_3^2 - 120 s_2^4 s_3^4 + 90 s_2 s_3^6 \\
- 8 s_2^8 s_4 - 56 s_2^5 s_3^2 s_4 + 428 s_2^2 s_3^4 s_4 + 16 s_2^6 s_4^2 \\
+ 64 s_2^3 s_3^2 s_4^2 - 112 s_3^4 s_4^2 - 128 s_2 s_3^2 s_4^3
\end{array}\right) \\[4pt]
&-16384 t^{14} s_3^2 \left(\begin{array}{c}
5 s_2^8 s_3^2 + 58 s_2^5 s_3^4 - 392 s_2^2 s_3^6 + 2 s_2^9 s_4 \\
+ 5 s_2^6 s_3^2 s_4 - 134 s_2^3 s_3^4 s_4 + 167 s_3^6 s_4 \\
- 16 s_2^7 s_4^2 - 104 s_2^4 s_3^2 s_4^2 + 528 s_2 s_3^4 s_4^2 \\
+ 32 s_2^5 s_4^3 + 32 s_2^2 s_3^2 s_4^3 - 64 s_3^2 s_4^4
\end{array}\right) \\[4pt]
&+256 t^{12} \left(\begin{array}
14 s_2^9 s_3^2 + 432 s_2^6 s_3^4 + 500 s_2^3 s_3^6 - 519 s_3^8 \\
+ 24 s_2^7 s_3^2 s_4 + 28 s_2^4 s_3^4 s_4 - 10832 s_2 s_3^6 s_4 \\
+ 16 s_2^8 s_4^2 - 64 s_2^5 s_3^2 s_4^2 + 1824 s_2^2 s_3^4 s_4^2 - 128 s_2^6 s_4^3 \\
- 1024 s_2^3 s_3^2 s_4^3 + 3072 s_3^4 s_4^3 + 256 s_2^4 s_4^4
\end{array}\right) \\[4pt]
&-32 t^{10} \left(\begin{array}{c}
94 s_2^7 s_3^2 + 2243 s_2^4 s_3^4 + 9328 s_2 s_3^6 + 36 s_2^8 s_4 + 744 s_2^5 s_3^2 s_4 \\
- 3648 s_2^2 s_3^4 s_4 - 144 s_2^6 s_4^2 - 1536 s_2^3 s_3^2 s_4^2
- 7680 s_3^4 s_4^2 + 2048 s_2 s_3^2 s_4^3
\end{array}\right) \\[4pt]
&+ t^8\left(\begin{array}{c}
81 s_2^8 + 1568 s_2^5 s_3^2 - 21184 s_2^2 s_3^4 + 768 s_2^6 s_4
+ 24064 s_2^3 s_3^2 s_4 \\
+ 37888 s_3^4 s_4 - 3072 s_2^4 s_4^2 - 24576 s_2 s_3^2 s_4^2\end{array}\right) \\[4pt]
&-4 t^6 (27 s_2^6 + 352 s_2^3 s_3^2 - 424 s_3^4 +
32 s_2^4 s_4 + 320 s_2 s_3^2 s_4 - 128 s_2^2 s_4^2) \\
&+ 2 t^4 s_2 (27 s_2^3 + 80 s_3^2) \\
&-12 t^2 s_2^2 \\
&+ 1
\end{align} \tag{$\star$}$$
Barring transcription errors, equation $(\star)$ gives an implicit relation between the area of a triangle and the lengths of its angle bisectors. (A numerical test against a random-ish GeoGebra model worked, so this can't be too far off.)
Now ... Where's that eternal fame I was promised?
If $e=f$, equation $(\star)$ reduces to
$$\begin{align}
0 &= ( 4 t - d f )( 4 t + d f ) \\
&\cdot (
256 t^6 d^4
+ 16t^4f^2 ( 9 d^6 + 4 d^4 f^2 + 4 d^2 f^4 + f^6 )
- t^2 d^2 f^6 ( 24 d^4 + 8 d^2 f^2 + 3 f^4 )
+ d^6 f^{10}
)^2 \\
&\cdot(
16 t^6 ( 4 d^2 - f^2 )
+ t^4 d^2 ( 64 d^4 - 32 d^2 f^2 + 9 f^4 )
- 2t^2 d^6 f^2 ( 8 d^2 + 3 f^2 )
+ d^{10} f^4 )
\end{align}$$
If $d=e=f$, then we have
$$(3 t^2 - d^4) (4 t - d^2)^3 (4 t + d^2)^3 (16t^4 + 19d^4t^2-d^8 )^3= 0$$
of which the first factor corresponds to the case of the equilateral triangle. The second and fourth factors yield positive real roots, so we aren't getting uniqueness from this thing.
Best Answer
We'll derive the equation using the fact: $$A_{PQR}=A_{ABC}-A_{PBR}-A_{RCQ}-A_{QAP}, \quad (I)$$ Using the angle bisector theorem we get: $$BP=\frac{ac}{a+b},\quad (1)$$ $$BR=\frac{ac}{b+c}, \quad (2)$$ $$CR=\frac{ab}{b+c},\quad (3)$$ $$CQ=\frac{ab}{a+c},\quad (4)$$ $$AQ=\frac{bc}{a+c},\quad (5)$$ and $$AP=\frac{bc}{a+b}. \quad (6)$$
Each mentioned area can be calculated using:
$$A_{PQR}=\frac{1}{2}ab\sin\gamma, \quad (7)$$ $$A_{PBR}=\frac{1}{2}BP\cdot BR\sin\beta, \quad (8)$$ $$A_{RCQ}=\frac{1}{2}CR\cdot CQ\sin\gamma, \quad (9)$$ and $$A_{QAP}=\frac{1}{2}AQ\cdot AP\sin\alpha. \quad (10)$$
Let $R$ be the circumradius, we know that: $$\sin \alpha = \frac{a}{2R}, \quad (11)$$ $$\sin \beta = \frac{b}{2R}, \quad (12)$$ $$\sin \gamma = \frac{c}{2R}, \quad (13)$$
Now if we substitute all the 13 equations in equation $(I)$ we get: $$A_{PQR}=\frac{1}{2} \cdot \frac{abc}{2R}-\frac{1}{2} \frac{a^2c^2b}{(a+b)(b+c)2R}-\frac{1}{2} \cdot \frac{a^2b^2c}{(b+c)(a+c)2R}-\frac{1}{2} \cdot \frac{b^2c^2a}{(a+b)(a+c)2R}, \Rightarrow$$
$$A_{PQR}=\frac{abc}{4R}[1-\frac{ac}{(a+b)(b+c)}-\frac{ab}{(b+c)(a+c)}-\frac{bc}{(a+b)(a+c)}], \Rightarrow$$
$$A_{PQR}=\frac{abc}{2R}[\frac{abc}{(a+b)(b+c)(a+c)}], \Rightarrow$$ $$A_{PQR}=A_{ABC}[\frac{2abc}{(a+b)(b+c)(a+c)}]$$ Using Heron's formula we are done.