The problem is as described in the title. In the following figure, we have two triangles $\triangle ABC$ and $\triangle DBE$, with some side-lengths and angle measures given. The goal is to find the area of the blue triangle:
The source of this problem is my good friend on Instagram @dare2solve who posts problems like these. They can also be found on the site that he listed in the picture above. I will post my approach to this problem as an answer below, please share your own approaches as well, and let me know if there is anything wrong with my approach or if my answer is incorrect!
Best Answer
I guess the problems you (and maybe your friend :)) propose, are (in general) supposed to be solved in an entirely geometric way. However, I am posting my solution which follows a very straightforward path.
Let's assume $\angle DBC=\theta$. Then,
$$\frac{\sin (115^{\circ}-\theta)}{\sin \theta}=\frac{BC}{DC}=\frac{DE}{DC}=\frac{\sin 115^{\circ}}{\sin 50^{\circ}}\implies \\\cos (65^{\circ}-\theta) -\cos(165^{\circ}-\theta)=\cos(115^{\circ}-\theta)-\cos(115^{\circ}+\theta)\implies \\-\cos(165^{\circ}-\theta)=\cos(115^{\circ}-\theta)\implies \\ \cos (15^{\circ}+\theta)=\cos(115^{\circ}-\theta)\implies \\ 15^{\circ}+\theta=115^{\circ}-\theta\implies \theta=50^{\circ};$$
and the rest is trivial.