Is this question solvable using the law of sines and cosines? 
*

*Is it possible to (analytically) calculate the area of the following triangle using the rules of sines and cosines?

*Is it possible to calculate it using only the rule of sines?

The data given is:
$BD$ is a median to $AC$, $\angle ABD = 50^\circ$, $\angle DBC = 18^\circ$ and $BD = 4.2~\text{cm}$.
It seems to me that some data is missing.
Am I right?

 A: Hint: extend $BD$ to $E$, such that $DE = BD$.  You should be able to find enough information in the triangle $\triangle BCE$ to calculate its area.
A: Let $x = AB$ , $ y = BC$, $\beta = \angle BDA $
Then
$\text{Area} = \dfrac{1}{2} (4.2) ( x \sin(50^\circ) + y \sin (18^\circ) ) = \dfrac{1}{2} x y \sin(68^\circ)\hspace{15pt} (1) $
And from the law of sines we have
$\dfrac{ x }{\sin \beta } = \dfrac{AD}{\sin(50^\circ)}$
$\dfrac{y}{\sin \beta } = \dfrac{DC}{\sin(18)^\circ}$
So by dividing these two equations we get
$ \dfrac{x}{y} = \dfrac{ \sin(18^\circ)}{ \sin(50^\circ) } \hspace{15pt} (2)$
Equation $(2)$ implies that
$ y = \left(\dfrac{ \sin(50^\circ)}{ \sin(18^\circ) } \right) x \hspace{15pt} (3)$
Substituting for $y$ from $(3)$ into $(1)$, and dividing by $x$ (because $x$ cannot be zero), results in
$ (4.2) ( 2 \sin(50^\circ) ) = x \dfrac{\sin(68^\circ) \sin(50^\circ) }{\sin(18^\circ)} $
Hence,
$ x = \dfrac{ 8.4 \sin(18^\circ) }{\sin(68^\circ)} $
and, consequently,
$ y = \dfrac{ 8.4 \sin(50^\circ) }{ \sin(68^\circ)}$
And finally, the area is
$\text{Area} = \dfrac{1}{2} x y \sin(68^\circ) = 35.28 \dfrac{ \sin(18^\circ) \sin(50^\circ) }{ \sin(68^\circ)} \approx 9.0074$
