This is a case of SAS. Could you do it without using the Laws of Sin.
My friend showed me how he could use the area formula for triangle 1/2 ac sinB for this. I don't understand. Could you attach a diagram?
[Math] Find the area of a rhombus ABCD with Side length of 3 and measure of angle ABC of 120.
geometrytrigonometry
Related Solutions
Note that from the law of sines you have all three equations $$\frac{\sin 120 ^\circ}{c}=\frac{\sin A}{10}=\frac{\sin B}{c-14}$$
Furthermore, you have that the angles $$A+B=60^\circ$$
This gives you enough equations. Remember to "convert" to radians before evaluating the sines.
EDIT: Actually, using the law of cosines is much easier, since (using $c=b+14$) you have $$(b+14)^2=10^2+b^2-2\times 10b\cos 120^\circ$$
However, from this one obtains $b<0$, which seems strange to say the least - the conclusion is therefore either that 1) the problem is flawed, 2) your reproduction of the problem is flawed, because the triangle cannot be constructed.
First, be careful with text book diagrams. I always tell my students that unless it is specified to be drawn to scale (which is very rare) then the only information that can be taken as true in a diagram are the actual measurements provided.
Having said that, you are correct. There are two possible angles for $\angle BDA$
My guess is that, since the Sine Law gives the smaller of the two possible angles, the text book writers were not necessarily expecting you to go further than this (are there any worked examples prior to this exercise that might give a clue as to the writer's expectations?). Either that, or they haven't provided enough info in the diagram to make one or the other answer the 'correct' answer, or, whoever wrote the answer booklet didn't go far enough.
Best Answer
Hint Draw the diagonals. They intersect each other perpendicularly, thus the result will be four congruent $30-60-90$ right triangles because $m \angle ABC = 120^\circ$.