If $ABCD$ is a cyclic quadrilateral with $AB=6, BC=4, CD=5, DA=3$ and $\angle ABC=\theta$, then $\cos \theta =$

The figure below represents the quadrilateral.

In triangle $ABC$, 

$A{C}^2=A{B}^2+B{C}^2-2\left(AB\right)\left(AC\right)\cos \theta$

$\Rightarrow A{C}^2=6^2+4^2-2\times 6\times 4\times \cos \theta$

$\Rightarrow A{C}^2=52-48\cos \theta ...\left(\mathrm{i}\right)$

Similarly, in triangle $ADC,$

$A{C}^2=A{D}^2+D{C}^2-2\times AD\times DC\cos \left(180°-\theta \right)$

$\Rightarrow A{C}^2=3^2+5^2+2\times 3\times 5\times \cos \theta$

$\Rightarrow A{C}^2=34+30\cos \theta ...\left(\mathrm{ii}\right)$

From equations $\left(\mathrm{i}\right) \& \left(\mathrm{ii}\right)$,

$52-48\cos \theta =34+30\cos \theta$$$

So, $\cos \theta =\frac{3}{13}$