A cattle feeding trough of uniform cross section and 2.5 m in length, is shown in Figure 1 - Leaving Cert Mathematics - Question 7 - 2019

For calculating $\angle{DOA}$, we can utilize trigonometric functions. Taking:

$\cos(\angle{DOA}) = \frac{60}{90} = \frac{2}{3}$
Therefore, $\angle{DOA} = \cos^{-1}(\frac{2}{3}) \approx 0.84 \text{ radians (correct to 2 decimal places)}.$

To obtain the area of the segment ABC, we calculate the area of the sector and then subtract the area of triangle AOC:

1. The area of the sector $OAC$: $\text{Area} = \frac{1}{2} r^2 \theta = \frac{1}{2} (0.9)^2 (\angle{DOA}) = \frac{1}{2} \cdot 0.81 \cdot 0.84 \text{ m}^2.$
2. The area of triangle AOC: $\text{Area} = \frac{1}{2} |OA| |OC| \sin(\angle{AOC}) = \frac{1}{2} \cdot 90 \cdot 90 \cdot \sin(\angle{AOC})$
   • where $\angle{AOC} = 180^\circ - \angle{DOA}.$
3. Therefore, the area of the segment ABC is: $\text{Area}_{ABC} = \text{Area}_{sector} - \text{Area}_{triangle} = 0.28 \text{ m}^2.$

To find the volume of the trough:

1. The volume $V$ is given by: $V = \text{Area}_{ABC} \times \text{Length} = 0.28 \text{ m}^2 \times 2.5 ext{ m} = 0.7 ext{ m}^3.$

The upper half consists of a hemisphere and a cylinder:

1. Volume of the hemisphere: $V_{hemisphere} = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (1.25)^3 = \frac{2}{3} \cdot \pi \times 1.953125.$
2. Volume of the cylinder: $V_{cylinder} = \pi r^2 h = \pi (1.25)^2 (3.5) = \pi \times 1.5625 \times 3.5.$
3. Adding these gives the total volume in the upper half.

To find the height $h$ of the remaining sand in the conical part, use the volume of the cone formula:

1. The volume for height $h$: $V_{cone} = \frac{1}{3}\pi r^{2} h = \frac{1}{3}\pi (1.25)^{2} h.$
2. Equating the volumes provides: $$$$

ightarrow h \approx 0.87 \text{ cm.}$$