Trigonometry (11th Edition) Clone

The volume of oil in the tank is $165~ft^3$
Let $C$ be the center of the circular end of the storage tank. We can form a triangle with two lines connecting $C$ with the two points where the top of the oil surface meets the side of the tank. Let's call these two points $A$ and $B$ and let $D$ be the midpoint of the horizontal line $AB$ The length of line $CD$ is $1~ft$ The length of lines $AC$ and $BC$ is $3~ft$ We can find the angle $\angle ACD$: $cos(\angle ACD) = \frac{1}{3}$ $\angle ACD = cos^{-1}(\frac{1}{3})$ $\angle ACD = 70.5^{\circ}$ The angle $\angle ACB = 2(70.5^{\circ}) = 141^{\circ}$ We can find the total area $A_1$ of the sector: $A_1 = (\frac{141^{\circ}}{360^{\circ}})(\pi r^2)$ $A_1 = (\frac{141^{\circ}}{360^{\circ}})(\pi) (3)^2$ $A_1 = 11.074~ft^2$ To find the area that the oil covers, we need to subtract the area of the triangle $\triangle ACB$. We can find the length of the side $AD$: $\overline{AD} = \sqrt{3^2-1^2} = \sqrt{8}$ We can find the area of $\triangle ACD$: $\frac{1}{2}(\sqrt{8})(1) = \sqrt{2}~ft^2$ Then the total area of the the triangle $\triangle ACB$ is $2\sqrt{2}~ft^2$ We can find the area covered by the oil: $A = 11.074~ft^2-2\sqrt{2}~ft^2 = 8.246~ft^2$ We can find the volume of the oil: $V = (8.246~ft^2)(20~ft) = 165~ft^3$ The volume of oil in the tank is $165~ft^3$