1. ## Chords and Arcs

2. Volume of the cylindrical tank=bh
$\displaystyle =\pi*(\frac{59}{2})^2*470$
$\displaystyle =1284314.95 in^3$

Volume of milk=$\displaystyle \frac{20}{59}*1284314.95 in^3 =435361 in^3$

$\displaystyle 231 in^3=1 gal$
$\displaystyle So, 435361 in^3=1885 gal$

3. Hello, juvenilepunk!

This requires some trig.

Look at the circular cross-section . . .

Code:
              * * *
*           *
*               *
*                 *

*         O         *
*         *         *
*      *  |θ * 29.5 *
*  9.5|     *
A*- - - - * - - - -*B
*:::::::C:::::::*
*:::::::::::*
* * *
D

The circle has radius $\displaystyle OA = OB = 29.5$ inches.
The depth of the milk is 20 inches, so $\displaystyle OC = 9.5$ inches.

We want the area of segment $\displaystyle ADB$
. . which is: .(area of sector $\displaystyle OADB$) - (area of triangle $\displaystyle OAB$).

In right triangle $\displaystyle OCB\!:\;\cos\theta \:=\:\frac{9.5}{29.5}\:=\:0.322033898$
. . Hence: .$\displaystyle \theta \:\approx\:71.214^o$
And: .$\displaystyle \angle AOB \:=\:2\theta \:\approx\:142.43^o$

$\displaystyle \text{Area of sector }OADB \:=\:\frac{142.43^o}{360^o}\times \pi(29.5^2) \:=\:1081.665251$ in².
$\displaystyle \text{Area of }\Delta AOB \:=\:\frac{1}{2}(29.5^2)\sin142.43^o \:=\:265.30887$ in².

Hence: .$\displaystyle \text{Area of segment }ADC \:=\:1081.665251 - 265.39887 \;=\;816.356381$ in².

The volume of the milk is: .$\displaystyle 816.356381 \times 470 \;=\;383687.4991$ in³.

Then: .$\displaystyle V \;=\;\frac{383687.4991\text{ in}^3}{1} \times \frac{1\text{ gal}}{231\text{ in}^3} \;=\;1660.984844 \;\approx\;1661\text{ gallons}$

4. AlexMahone's answer is incorrect be 20/59 of the diameter would be 20/59 of the volume only if the volume were a linear function of the diameter- and it isn't.

5. ^Sorry.

Soroban is right.