I have a triangle of $600m^2 $ area, I do not have any lengths of this triangle, I only have the rise/run of both sides that are not sitting on the x-axis, and the area, can I find the height given the relation it has with the angles of the triangle?
Angles of this triangle in degrees are $23.94, 23.12, 132.94$ I calculated this by
calculating $\tan^{-1}$ of the both slopes (rise/run)
slope1=$0.444$
slope2=$0.429$
(image provided is a rough similarity)
How would I approach this question?
The area is $$\mathcal{A}=\frac{1}{2}CD(AD+DB).$$ We know the slopes, the angles are not necessary. It suffices to realize that the slopes are $$\frac{CD}{AD}\quad\text{and}\quad\frac{CD}{DB},$$ respectively. Then $$AD=0.444\times CD, DB=0.429\times CD.$$ Putting this and the known value of the area into the first equation you obtain the height.