This should be easy! Right? Part - 2

$\angle ADB=Cos^{-1}\frac 7 8=\alpha=Sin^{-1}\frac{\sqrt{15}} 8=\angle BDC=\angle ACB. ~\therefore ~ABCD ~cyclic.\\ \therefore ~\angle CAB=\alpha, ~ \implies ~BC=6. ~~~ Let ~\angle DCA=Cos^{-1}\frac{11}{16}= say ~ \beta.\\ SinCBD=Sin(2*\alpha+\beta). ~~\\ Applying ~Sin ~ Law, ~\dfrac{DC}{Sin\beta}=\dfrac 6 {Sin\alpha}.\\ Putting ~ the ~ Values, CD=12 cm.$

We know,

$\cos^{2} \angle ACB = 1 - \sin^{2} \angle ACB$

$\cos^{2} \angle ACB = 1 - \frac{15}{64}$

$\cos^{2} \angle ACB = \frac{49}{64}$

$\cos \angle ACB = \frac{7}{8}$

Notice $\cos \angle ACB = \cos \angle ADB$ ,

Hence,

$\angle ACB = \angle ADB$

It is sufficient to prove that quadrilateral $ABCD$ is cyclic.

After that we can easily apply $cosine$ rule,etc.