finding Length of a diagonal Given Quadrilateral ABCD in such that $AB<BC<CD$ creating increasing arithmetic progression with sum of $27$ cm.
$\measuredangle BCD=60^{0}$. the diagonal $BD=\sqrt{133}$ cm, and it divided $\measuredangle ABC$ such that:
$\measuredangle CBD=2\measuredangle DBA $ .  compute the length of diagonal $AC$.
This is a bit strange because computing by law of sines or law of cosines will give different number of solutions and i can't really prove that one of the solutions is disqualified.
If we look at triangle $BCD$: $\frac{\sqrt{133}}{sin60^{0}}=\frac{9}{sin\measuredangle BDC}$
so $\measuredangle BDC=42.5^{0}$ and $\measuredangle DBC=77.48^{0}$
But if i compute: $\frac{\sqrt{133}}{sin60^{0}}=\frac{13}{sin\measuredangle DBC}$  so  $\measuredangle DBC=77.48^{0}$ or $102.51^{0}$  
 A: Let $AB=x$, $BC=x+y$, $CD=x+2y$. The sum is $27$, so $x+y=9$.
Now use the Cosine Law. We have $CD=9+y$, and therefore
$$133=(9+y)^2+81-(2)(9)(9+y)(1/2).$$
This simplifies to $y^2+18y-52=(y+13)(y-4)=0$, so $y=4$. Thus our arithmetic progression is $5,9,13$.
Now we can use the Cosine Law again to find the cosine of $\angle CBD$, and then trigonometric identities to find the cosine of $\angle ABC$, and then the Cosine Law to find $AC$. Although exact expressions can be found, they are not particularly attractive. The angle $ABC$ turns out to be approximately $116.22$ degrees.
Remark: Using the Sine Law is less efficient: the cosine identifies the angle of a triangle uniquely, but the sine does not.  So using the Sine Law to compute $\angle DBC$ leads to extra work.  There is only one possible answer, since the sides determine the angles.
We can rule out the possibility that $\angle DBC$ is obtuse in various ways. For example, we can compute the sine of $\angle CDB$, the smallest angle, by using the Sine Law. There will be no ambiguity, the angle turns out to be about $42.52^\circ$. Now the remaining angle $DBC$ gets uniquely identified: it is $\approx 180-(60+42.52)$.
