Here is a very similar task in this forum which was solved brilliantly by the user Apfelkuchen.
The key to the right solution is a correct graphing an equation and the notion of derivative of function. Let's assume that the point from where the surveying team determines the height of the hill is the point of inter-crossing of two lines of linear graphs of functions f(x) = tan x and g(x) = tan (x - 3.658) (for the convenience sake I convert feet into meters: 12 feet = 3.658 meters). Let's place the graphs as shown in the fig.below, where A - is the determination point, AC - the hill's root, B - its top and at the same time the bottom of the pole OB with O being the pole's top, x = OC is the coordinate on the Ox axis of the inter-crossing point A. O has coordinates (0 ; 0), while B has coordinates (3.658 ; 0).
GRAPH.png
It's easy to calculate that the angle AOB = 14 grades, the angle ABC = 20 grades, so the linear graph ax, where a is the slope coefficient of the blue line, has derivative that is presented here by the line's extreme position and is equal to tan (14 grades) = 0.249. The linear graph b(x - 3.658) is the extreme position of the green line and is determined by the slope coefficient b = tan (20 grades) = 0.364. Considering these presumptions and the fact of the two having the common point A, we now can find its x coordinate by composing the equation tan 14x = tan 20(x - 3.658) => x = [tan 20*3.658]/[tan 20 - tan 14] = [0.364*3.658]/[0.364 - 0.249] = 11.578365 ~ 11.578 meters, from which the height of the hill is ~ 11.578 - 3.658 ~ 7.92 meters.
+259 
