The diagram shows a semicircle, centre O, radius 16 cm.

1. Find the distance from $O$ to the foot of the perpendicular on chord $AB$.

Let $C$ be the point on the diameter where the perpendicular from $O$ meets chord $AB$.

The radius is $16$ cm and the labelled gap to the end of the diameter is $3.2$ cm, so

$OC = 16 - 3.2 = 12.8\text{ cm}.$

2. Use cosine to find half of the central angle, then double it.

Since $OA=OB$, the perpendicular from $O$ to chord $AB$ bisects $\angle AOB$.

In right-angled triangle $OCB$,

$\cos \angle COB = \frac{OC}{OB}=\frac{12.8}{16}=0.8.$

$\angle COB = \cos^{-1}(0.8)=0.6435\ldots\text{ radians}.$

Hence

$\angle AOB = 2(0.6435\ldots)=1.2870\ldots\text{ radians}.$

1. Find the area of sector $AOB$.

Using $r=16$ cm and $\theta=1.287$ radians,
$\text{Area of sector }AOB=\frac12 r^2\theta$
$=\frac12(16)^2(1.287)=164.736\text{ cm}^2$

2. Find the area of triangle $AOB$.

Since $OA=OB=16$ cm and the included angle is $1.287$ radians,
$\text{Area of }\triangle AOB=\frac12 ab\sin C$
$=\frac12(16)(16)\sin(1.287)$
$=128\sin(1.287)\approx 122.880\text{ cm}^2$

3. Subtract the triangle area from the sector area to get the shaded segment area.

$\text{Shaded area}=\text{sector area}-\text{triangle area}$
$=164.736-122.880$
$=41.856\text{ cm}^2$
$\approx 41.9\text{ cm}^2$

1. Convert the length of the trough from metres to centimetres.

$2.8\text{ m}=2.8\times100=280\text{ cm}$

2. Multiply the cross-sectional area by the length of the trough.

Using the shaded area from part (b),
$\text{Volume}=41.9\times280=11732\text{ cm}^3$

3. Write the volume in standard form.

$11732=1.1732\times10^4$
To 3 significant figures,
$11732\approx1.17\times10^4$

1. Calculate the total capacity of the trough in cubic centimetres.

The cross-section of the full trough is a semicircle of radius $16$ cm.

Length of trough $=2.8\text{ m}=280\text{ cm}$.

$\text{Capacity}=\frac12\pi(16)^2\times 280$

$=35840\pi=112594.7\text{ cm}^3$

2. Subtract the volume of water already in the trough.

From part (c)(i), the volume of water already in the trough is

$41.9\times 280=11732\text{ cm}^3$

So the volume still needed is

$112594.7-11732=100862.7\text{ cm}^3$

3. Convert the remaining volume from cubic centimetres to litres.

Since $1000\text{ cm}^3=1$ litre,

$100862.7\div 1000=100.8627\text{ litres}$

$\approx 101\text{ litres}$