Upper figure: cylinder oil tank cross-section perpendicular to its horizontal axe. The vertical coordinate is the oil level in percentage.
Lower figure: graph of oil volume/max. volume (in %) versus oil level $l$ (in feet). The horizontal straight lines represent the area/volume ratio $A(l)/A(10)=V(l)/V(10)$ (in %) for every multiple of 10% from 0% to 100%.
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Since the tank radius is $5$, the oil level with respect to the bottom of
the tank is given by $l=5-5\cos \frac{\theta }{2}$, where $\theta $ is the
central angle as shown in the figure. The area of the tank cross section
filled with oil is
$$A(\theta )=\frac{25}{2}\theta -\frac{25}{2}\sin \theta $$
or
$$A(l)=25\arccos (\frac{5-l}{5})-\frac{25}{2}\sin (2\arccos (\frac{5-l}{5}))$$
The area ratio $A(l)/A(10)=V(l)/V(10)$ where $V(l)$ is the oil volume.
Let $f(l)$ denote this area ratio in percentage:
$$f(l)=\frac{100}{\pi }\arccos \left( 1-\frac{1}{5}l\right) -\frac{50}{\pi }\sin \left( 2\arccos \left( 1-\frac{1}{5}l\right) \right) $$
Here is the sequence of $f(l)$ values for $l=0,1,2,\ldots ,10$. The graph of $f(l)$ is shown above.
$f(0)=0$, $f(1)=5.2044$, $f(2)=14.238$, $f(3)=25.232$, $f(4)=37.353$, $f(5)=50$,
$f(6)=62.647$, $f(7)=74.768$, $f(8)=85.762$, $f(9)=94.796$, $f(10)=100$
Edit: One still needs to solve the nonlinear equation $f(l)-10k=0$ for $k=1,2,3,4,6,7,8,9$, e. g. by the Secant Method.
Edit 2: The problem of solving graphically, as shown in the secong figure, is that it would be very difficult, or impossible, to get the required accuracy of 0.01 (feet).
Update: The oil level marks (in feet) should be placed at
$0,1.57,2.54,3.40,4.21,$
$5,5.79,6.60,7.46,8.44,10$
corresponding to the oil volume percentage of
$0,10,20,30,40,$
$50,60,70,80,90,100$.
This calculation was based on the following $f$ function values:
$f(0)=0.0$, $f(1.5648)=10.0$, $f(2.5407)=20.0$, $f(3.40155)=30.0$, $f(4.21135)=40.0$
$f(5)=50.0$, $f(5.7887)=60.0$, $f(6.59845)=70.000$, $f(7.4593)=80.000$,
$f(8.4352)=90.000$, $f(10)=100.0$
Update 2 Figure of marks:
[Rearranged to show the sequence of editions and updates.]
I always think of this sort of stuff in terms of forcing the the result to have the units you want and/or the only units that can possibly make sense.
In the first case, you are given $\frac{rotations}{minutes}$ and $feet=\frac{feet}{1}$ and you want to end up with velocity, which has the general form $\frac{distance}{time}$. In this case, the pertinent units for distance and time are feet and minutes, respectively. The key, though, is the following: one rotation equals $2\pi$ times the radius (in feet). That is, $$\frac{rotations}{minutes}=2\pi\cdot radius\cdot\frac{feet}{minutes}.$$
But remember that in the end we just want an expression for velocity; that is, $\frac{feet}{minutes}=$ something. That's simple, though. Just divide both sides of the above equation by $2\pi\cdot radius$.
The second and third ones are very similar, but angular velocity has units $\frac{rotations}{minutes}$.
The last thing you asked is a specific instance of this sort of dimensional analysis, except that you have to convert everything to miles in the end. Alternatively, you can think of it as follows: if the velocity is 50 feet/second and the angular speed is 100 revolutions/second, that means in one second, the apparatus is making 100 revolutions - and furthermore, that these 100 revolutions = 50 feet. From this you get that one revolution is 1/2 a foot. That means that the circumference of the circle drawn out by the motion of the apparatus is 1/2 a foot, meaning that the radius of the circle is $\frac{1}{2}\cdot\frac{1}{2\pi}$ feet. To turn this into miles, just divide by 5280. Hope that helps.
Best Answer
In your answer, you seem to be confused about the angle. The figure looks like this (I'm not drawing to scale, by intention.)
The tourist is at $A$, the friend $C$, and the problem tells you $\beta=85.6^\circ$. The key insights are then (1) clarifying whether "distance" refers to the direct fly distance AC or the ground distance BC; this is not clear to me from your post (2) recognizing that $\alpha=\beta$ and then invoking the appropriate trig relation based on your definition of "distance".
If the tourist gets down to the bottom of the observation tower, then travels on ground to the friend, then "distance" $d$ ($=BC$) satisfies $$ 520/d = \tan(\alpha)=\tan(85.6^\circ)\implies d=40.0118. $$ If the tourist can fly directly to the friend, then "distance" $d$ ($=AC$) satisfies $$ 520/d=\sin(\alpha)=\sin(85.6^\circ)\implies d=521.537. $$ Given the answer, the teacher intended the latter but, in my opinion, the question is ambiguous.