#Cyclic quadrilateral

Let the side lengths of quadrilateral $ABCD$ be $a=3$ , $b=4$ , $c=5$ , and $d=6$ . By Ptolemy's theorem , the diagonal $AC$ :

$q = \sqrt{\frac {(ac+bd)(ad+bc)}{ab+cd}} = \sqrt {\frac {247}7}$

The diameter $AE$ of a circumcircle is given by

$D = \frac {AC}{\sin B} = \frac q{\sin B} \implies \sin B = \frac qD$

Since $AE$ is a diameter, by Thales theorem we have $\angle ACE = 90^\circ$ . Let $\angle CAE = \theta$ . Then

$\begin{aligned} \sin \theta & = \frac {CE}{AE} = \frac {\sqrt{AE^2-AC^2}}{AE} \\ & = \frac {\sqrt{D^2-q^2}}D = \sqrt{1-\left(\frac qD\right)^2} \\ & = \sqrt{1-\sin^2 B} = \blue - \cos B & \small \blue{\text{Since }\angle B \text{ is obtuse,}} \end{aligned}$

By cosine rule ,

$\begin{aligned} q^2 & = a^2 + b^2 - 2ab \cos B \\ \implies \cos B & = \frac {a^2 + b^2 - q^2}{2ab} = \frac {7(3^2+4^2)-247}{2\cdot 3 \cdot 4 \cdot 7} = - \frac 37 \\ \implies \sin \theta & = - \cos B = \frac 37 \end{aligned}$

Therefore $m+n = 3+7 = \boxed {10}$ .

Draw the center $O$ and $\triangle AOC$ :

By the properties of a cyclic quadrilateral , $\cos \angle D = \cfrac{a^2 + d^2 - b^2 - c^2}{2(ad + bc)} = \cfrac{5^2 + 6^2 - 3^2 - 4^2}{2(5 \cdot 6 + 3 \cdot 4)} = \cfrac{3}{7}$ .

As a central angle intercepting the same arc of inscribed angle $\angle D$ , $\angle AOC = 2\angle D$ .

As a base angle of isosceles triangle $\triangle AOC$ , $\angle CAE = \frac{1}{2}(180° - \angle AOC) = 90° - \angle D$ .

Therefore, $\sin \angle CAE = \sin (90° - \angle D) = \cos \angle D = \cfrac{3}{7}$ , so $m = 3$ , $n = 7$ , and $m + n = \boxed{10}$ .