Figure A is the right way to go about this. I just drew a perpendicular from the angle 105 to the base. This will give you one 30-60-90 triangle and one 45-45-90 triangle. You can easily find the area then
Figure B is the figure that goes by your approach, firstly this is not a 30-60-90 triangle you see. The triangle with the perpendicular has angles 15-90-75 which don’t have any side ratios like 30-60-90 or 45-45-90 unless Trigonometry is used so because we have the side ratios present for figure A approach that is a better way to solve it.
But how does one know that by drawing a perpendicular line segment from the topmost line to its opposite vertex (with the 105’ angle), we can split 105’ into 45’ and 60’ angles exactly (and not some other angles)? Seems too convenient?
Okay, never mind. I figured it out. The moment one draws a perpendicular line segment, you introduce two 90’ angles and two triangles. This means 105’ must be split into 45’ and 60’.
Just to clarify, we identified that it would be divided into 45 and 60 because we were given 30 degrees. So, using 30 and 90 in one triangle, I found out 60 and then just subtracted it from 105 to get 45. It might not always be the case that they would be divided in this ratio. Because 30 was given there, i could find these ratios.