I've found myself trying to solve this for my Geometry class where we have to model a basic piece of architecture and find its volume and surface area (very basic). But the structure I chose requires me to find its dimensions via some other given dimensions. I've made a diagram linked below. So what is the altitude (and how did you find it, if possible?)
The altitude of your triangle is 24. You can conclude this by analytic geometry. Let's work with the xy-plane. Suppose that the $x$-axis contains the bottom side of your triangle and that the $y$-axis contains your red vertical segment. If the line $L$ that contains the right side of the triangle has equation $y=mx+c$, then $c$ is the altitude of the triangle.
Lets denote by $D$ the red diagonal (segment) of length 6 of your diagram. Choose $D$ such that it's perpendicular to the right side of the triangle. Since the angle between the $y$-axis and $D$ is 60, the line that contains $D$ has equation $y=\frac{\sqrt{3}}{3}x+12$. Using this equation we can deduce that $m=-\frac{3}{\sqrt{3}}$ and that the intersection of $D$ and the right side of the triangle is the point $(3\sqrt{3},15)$. Since this point is part of $L$ we conclude that $L$ has equation $y=-\frac{3}{\sqrt{3}}x+24$. Hence $c=24$.