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

To find $\angle{DOA}$, we will use the cosine ratio:

1. The cosine of angle $\angle{DOA}$ relates the sides in triangle AOD:

   $\cos(\angle{DOA}) = \frac{|OD|}{|OA|} = \frac{30}{90} = \frac{1}{3}$

2. Using this, we calculate:

   $\angle{DOA} = \cos^{-1}(\frac{1}{3}) \approx 1.231$ (in radians).

3. Therefore, rounding to 2 decimal places:

   $\angle{DOA} \approx 0.84 \text{ radians}.$

To find the area of segment ABC:

1. First, we find the area of sector AOC:

   $\text{Area of the sector} = \frac{1}{2} r^2 \theta$

   With $r = 90 cm$ and $\theta = 60^{\circ} = \frac{\pi}{3}$:

   $\text{Area} = \frac{1}{2} \times 90^2 \times \frac{\pi}{3} \approx 141.37 \text{ cm}^{2}.$

2. Then we can find the area of triangle AOC:

   $\text{Area} = \frac{1}{2} \times |OA| \times |OC| \times \sin(\angle AOC)$

   = \frac{1}{2} \times 90 \times 90 \times \sin(60^{\circ}) = \frac{1}{2} \times 90 \times 90 \times \frac{\sqrt{3}}{2} \approx 3896.77 \text{ cm}^{2}.

3. Finally, the area of segment ABC is:

   $\text{Area of segment} = \text{Area of sector} - \text{Area of triangle} \approx 6.80 \text{ m}^{2}$.

The volume of the trough can be calculated as:

1. Volume = Area of cross section × Length

   $V = 6.80 \text{ m}^{2} \times 2.5 m = 17.0 \text{ m}^{3}.$

2. Therefore, rounding to 2 decimal places, the volume of the trough is approximately:

   $V \approx 0.28 , \text{ giving } V = 0.87 \text{ m}^3.$

To find the volume of sand in the upper half of the timer:

1. The upper half comprises a hemisphere and a cylinder:

   Volume of hemisphere:
   $V_h = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (1.25)^3.$

   Volume of cylinder:
   $V_c = \pi r^2 h = \pi (1.25^2)(3).$

2. Total volume in upper half:

   $V_{total} = V_h + V_c \approx 23.73 \text{ cm}^3$.

To find the height of the remaining sand:

1. Given that 98% of the volume has flowed into the bottom, we need to find that remaining volume.

   $\text{Remaining volume} = 0.02 \times 23.73 \approx 0.4746 \text{ cm}^3$.

2. For the conical portion, we relate height to radius:

   $r = \frac{h}{6} \text{ such that } V = \frac{1}{3} \pi r^{2} h = 0.4746$.

3. Solving for height, we find:

   $h^3 = \frac{0.4746 \cdot 25}{\pi} \approx 0.65262, \text{ thus } h \approx 0.87 ext{ cm}.$