Strange cyclic quadrilateral

By the (extended) Sine Rule, applied to appropriate triangles (all of which have the same circumcircle as the cyclic quadrilateral):

$\frac{a}{\sin\alpha} \; =\; \frac{b}{\sin\beta} \; = \;\frac{c}{\sin\theta} \; = \; \frac{d}{\sin\gamma} \; = \; 2R$

where $R$ is the radius of the circumcircle of the cyclic quadrilateral. Thus $X \; = \; \sin\alpha + \sin\beta + \sin\gamma + \sin\theta \; =\; \frac{a+b+c+d}{2R}$ Now $a,b,c,d$ are the zeros of the polynomial $f(X) \; =\; X^4 - 24X^3 + 201X^2 - 698X + 844 \;,$ and hence $a+b+c+d = 24$ , so that $X = \tfrac{12}{R}$ .

The circumradius of a cyclic quadrilateral with sides $a,b,c,d$ is

$R \; = \; \frac14\sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)}}$

where $s = \tfrac12(a+b+c+d)$ is the semiperimeter. But $\begin{array}{rcl} a^2+b^2+c^2+d^2 & = & (a+b+c+d)^2 - 2(ab + ac + ad + bc + bd + cd) \\ & = & 24^2 - 2\times201 \; = \; 174 \\ a^2b^2c^2 + a^2b^2d^2 + a^2c^2d^2 + b^2c^2d^2 & = & (abc + abd + acd + bcd)^2 - 2abcd(ab+ac+ad+bc+bd+cd) \\ & = & 698^2 - 2\times844\times201 \; = \; 147916 \\ (ab+cd)(ac+bd)(ad+bc) & = & abcd(a^2 + b^2 + c^2 + d^2) + (a^2b^2c^2 + a^2b^2d^2 + a^2c^2d^2 + b^2c^2d^2) \\ & = & 844 \times 174 + 147916 \; = \; 294772 \\ (s-a)(s-b)(s-c)(s-d) & = & f(s) \; = \; f(12) \; =\;676 \end{array}$ and hence $R \; = \; \frac14\sqrt{\frac{294772}{676}} \; = \; \frac{\sqrt{73693}}{52}$ and hence $X \; = \; \frac{12\times52}{\sqrt{73693}} \;=\; \frac{624}{\sqrt{73693}}$ making the answer $\boxed{74317}$ .