Lets bring this to a whole new level

This is tagged in geometry. I would solve it with geometry.

Considering the graph of $\sin{(x°)}$ , it can be seen that the maximum occurs when $n=180$ since from then on, it becomes negative. So, in the function $\sin{(x°)}$ , there are $180$ terms between $0\le x°\le 180°$ or $0\le x°\le\pi$ . With so many terms squished in such a small "space", we can estimate this summation to integration.

To find an estimate of the average of all the terms, we can integrate the function: $\int _{ 0 }^{ \pi }{ \sin ^{ 3 }{ (x) } dx } =\frac { 4 }{ 3 }$ This gives us the area underneath the graph between $0\le x°\le\pi$ . Then, we can assume this area to be of a rectangle with its base length as $\pi$ . So, dividing the above value by $\pi$ would give the estimate of the average of all the terms: $\frac { \left( \frac { 4 }{ 3 } \right) }{ \pi } =\frac{4}{3\pi}$ Finally, to find an estimate of the answer, we just have to take that and times the number of terms:

$\frac{4}{3\pi}\times180=76.39437268410976...$ This is very very close to the exact answer: $76.3943727727258...$ Therefore, rounding up to the nearest integer would give $\boxed{76}$

$\textbf{NOTE:}$ This is fairly accurate because a lot of terms are squished into a small space, making it similar to integration.

Feel free to post solutions that would give a more accurate answer.

Amazing problem, my approach is quite different,

We can obviously write the given expression as

$\sum _{ i=1 }^{ n }{ (\frac { 3sin(\frac { 2\pi k }{ 180 } ) }{ 4 } } \quad -\quad \frac { sin(\frac { 2\pi k }{ 120 } ) }{ 4 } )$

Now this is simply the partial sum (depends on value of n ofcourse) of the complex roots of unity, one set is of the the 360th and the other 120th roots of unity,

Clear we would want one to be maximum (the one with positive sign ) the first one and the other with negative sign to be minimum,

The first one will be maximum at the peak point or n= 180 k for some odd integer k , (as the sum is 0 when all roots are added and max at the middle)

But the second one is never 0 for any 180k where k is odd number,

so we see that it atleast with holds to minimum value possible,

at k=1, we will make second term maximum (sum uptill 120 + 60 and 60 is middle term)

at k=3, is when we get 76 as the closest integer, we can see that again the sum falls above,

i agree that this is a bit of a trial and error but also in this simple case, i found it simpler this way than to use the formula for sum of the roots of unity and find maxima