Problem: Oil Tank Problem
We want to determine if a certain Superintendent's cylindrical tanks has a significant amount of oil to finish the winter season. To explore this problem, it suffices to create a diagram with the given information using GSP to begin an estimation.
Notice that the angle bisector of angle CBA creates two congruent right triangles. Triangles CBD and ABD are congruent by SAS. Finding the angle measure of triangle ABD and lenght of segment AD will jump start our estimation of the amount of oil in the tank. We know two measurements of the right triangle ABD, the hypotenuse and the adjacent side of angle DBA. So we can use a trigonometric ratio to the measure of angle DBA.
So we have,
cos (theta/2) = 8/18
(theta) = 2 cos^-1(8/18)
theta = 2.22048
The length of segment AD:
a^2 + b^2 = c^2
8^2 + b^2 = 18^2
64 + b^2 = 324
b^2 = 260
b = 16.1245
Since triangles CBD and ABD are congruent, the length of segment CD is also 16.1245 inches long. We have found the missing parts of the triangles, and now we must find the areas of the sector and small triangles, which will help yield the area of the oil in the tank.
Area of the sector: A = (1/2)(r^2)(theta) = [(1/2)(324)(2.22048)] = 357.178
Area of small triangles: [16.1245(8)(1/2)] = 128.996
To find the area of the oil in the tank, it suffices to determine the difference of the area of the sector and the area of the triangles.
Area of the oil in tank = Area of the sector - Area of the triangle
Area of the oil in tank = 357.178 - 128.996 = 228.182
Volume = 48( 228.182) = 10952.7 cubic inches
Gallons of Oil : (10952.7)/(231) = 47.4144
So the superintendent has enough oil to finish the season.
General Case
Volume of any given radius, r, and height of oil in tank, n
Volume = 48 ( (324 cos^-1 (18-n/18) -((r-n)(r^2-(r-n)^2)^1/2)