please help with trig question!

I  think  there  might  be  some "formula" for solving this in an easier fashion, but, I believe  the Law of Cosines  might also work

Note  that angles  DCB  and DAB  are supplemental   so     cos ( DCB)   =  -cos (DAB)

Using the Law of  Cosines we have this system    (and subbing  -cos (DAB) for cos (DCB)  )

DB^2   =  3^2  + 2^2   - 2 ( 3 * 2)  (-cos DAB)

DB^2   =  1^2   + 4^2  - 2 ( 4 * 1)   (cos DAB)

DB^2  =  13  +  12 cos DAB

DB^2  =  17   -   8 cos DAB

Multiply  the  first equation  by     8    and the  second  by    12

8 DB^2   =  104  + 96  cos DAB

12 DB^2  = 204  -  96  cos DAB               add  these

20 DB^2  =   308

DB^2  =  308  / 20     =    15.4  

The side lengths of a cyclic quadrilateral ABCD are provided in the diagram. Find BD^2.

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a = 1     b = 2     c = 3     d = 4      s = 5

Using the above formula, we can calculate the circumradius R.

R = 2.002602473

Now, we can calculate the measure of angle C.

Angle C = 101.53695902º

Finally, using the law of cosines, we can find the length of BD.

BD ≈ 3.924         BD² = 15.4