Let me demonstrate my inner autist a bit:
First let's call the point at the bottom middle at the right end of the known 10cm "S" and the inner line "t", while the remaining sides get the default labels of a, b, c to the opposite of each corner A, B, C each, and la and ra for the left and right parts of a.
We know that every Triangle has a total (inner) angle of 180°, so for ACS at S, the angle is 180°-15°-15°=150°.
And because we obviously know that the angle of S for ABC is 180°, we can deduce the angle of S for ABS is 180°-150°=30°, which in turns let's us deduce that the angle in ABS at A is 180°-15°-30°=135°, with the angle of ABC at A therefore being 135°+15°=150° (though this last part is redundant).
We have all the "easy" info now, so we can now start calculating a bit using the law of sines:
Now applying Heron's formula:
So the area of ABC is 12.47 cm^2 +/- rounding errors.
I am sure someone will soon tell me how overcomplicated I made this and that there is supposedly a much simpler solution, but eh, whatever.