Answer: 47

Solution:

Label the quadrilateral $ABCD$ with $AB = 5,$ $BC = 6,$ $CD = 9,$ $DA = 7,$ and let the diagonals meet at $P,$ cutting $\overline{AC}$ into $p_1, p_2$ and $\overline{BD}$ into $q_1, q_2.$ With $\varphi = \angle APB,$ the law of cosines in the four corner triangles (whose angles at $P$ alternate between $\varphi$ and $180^\circ - \varphi$) gives $$\begin{aligned} &BC^2 + DA^2 - AB^2 - CD^2 \\ &= 2(p_1q_1 + p_1q_2 + p_2q_1 + p_2q_2) \\ &\quad {}\cdot \cos\varphi \\ &= 2\,AC \cdot BD \cos\varphi. \end{aligned}$$

The left side is $36 + 49 - 25 - 81 = -21,$ so $AC \cdot BD\,|\cos\varphi| = \frac{21}{2},$ and the acute angle $\theta$ between the diagonals satisfies $AC \cdot BD \cos\theta = \frac{21}{2}.$ Meanwhile the four corner triangles give the area $\frac{1}{2} AC \cdot BD \sin\theta = 30,$ so $AC \cdot BD \sin\theta = 60.$

Dividing, $\tan\theta = \frac{60}{21/2} = \frac{40}{7},$ so $m + n = 40 + 7 = 47.$