1. **Problem Statement:** Calculate the total volume of a water tank composed of a cylinder and a cone on top. Given: - Diameter of cylinder EF = 4 ft - Height of cylinder BF = 5 ft - Angle $ m\angle ABD = 71.6^\circ $ 2. **Identify dimensions:** - Radius of cylinder and cone base $ r = \frac{EF}{2} = \frac{4}{2} = 2 $ ft - Height of cylinder $ h_{cyl} = BF = 5 $ ft 3. **Find height of cone $ h_{cone} $:** - The angle $ \angle ABD = 71.6^\circ $ is between the vertical line AB and the slant height AD of the cone. - Using right triangle ABD, where AB is vertical height of cone, BD is radius (2 ft), and AD is slant height. - $ \tan(71.6^\circ) = \frac{BD}{AB} = \frac{2}{h_{cone}} \Rightarrow h_{cone} = \frac{2}{\tan(71.6^\circ)} $ Calculate $ h_{cone} $: $$ h_{cone} = \frac{2}{\tan(71.6^\circ)} \approx \frac{2}{2.967} \approx 0.674 \text{ ft} $$ 4. **Volume formulas:** - Cylinder volume: $$ V_{cyl} = \pi r^2 h_{cyl} $$ - Cone volume: $$ V_{cone} = \frac{1}{3} \pi r^2 h_{cone} $$ 5. **Calculate volumes:** - Cylinder: $$ V_{cyl} = \pi \times 2^2 \times 5 = 20\pi $$ - Cone: $$ V_{cone} = \frac{1}{3} \pi \times 2^2 \times 0.674 = \frac{1}{3} \pi \times 4 \times 0.674 = \frac{4}{3} \pi \times 0.674 = 0.899\pi $$ 6. **Total volume:** $$ V_{total} = V_{cyl} + V_{cone} = 20\pi + 0.899\pi = 20.899\pi $$ 7. **Numerical approximation:** $$ V_{total} \approx 20.899 \times 3.1416 = 65.63 \text{ cubic feet} $$ 8. **Final answer:** The total volume of the tank is approximately **66 cubic feet** to the nearest cubic foot.