We shall consider the graph of the $\sin$ function.

Since the values of $\sin t^{\circ}$ from $180^{\circ}<t^{\circ}<360^{\circ}$ are negative, $\sin t^{\circ} = \sin(180-t)^{\circ}$ for all $t$ such that $0^{\circ} \leq t^{\circ} \leq 180^{\circ}$ and the period of the $\sin$ function is $360^{\circ}$ , it suffices to use $n=180$ .

Now, we want to find a formula for $\sin t^{\circ}+\sin (2t)^{\circ}+\cdots+\sin (nt)^{\circ}$ . We proceed to multiply and divide the expression by $2\sin\left(\frac{t}{2}\right)^{\circ}$ to obtain

$\frac{2\sin\left(\frac{t}{2}\right)^{\circ}\left[\sin t^{\circ}+\sin (2t)^{\circ}+\cdots+\sin (nt)^{\circ}\right]}{2\sin\left(\frac{t}{2}\right)^{\circ}}$

We next distribute the $2\sin\left(\frac{t}{2}\right)^{\circ}$ to all the terms in the numerator. We also know that $2\sin x^{\circ}\sin y^{\circ}=\cos(x-y)^{\circ}-\cos(x+y)^{\circ}$ . Hence, by letting $x=\frac{t}{2}$ and $y=t$ , we obtain a telescopic sum which cancels out to be

$\sin t^{\circ}+\sin (2t)^{\circ}+\cdots+\sin (nt)^{\circ}=\frac{\cos\left(\frac{t}{2}\right)^{\circ}-\cos\left(\frac{(2n-1)t}{2}\right)^{\circ}}{2\sin\left(\frac{t}{2}\right)^{\circ}}$

By letting $t=1$ and $n=180$ , we have our desired answer to be $\boxed{115}$ , and we are done. $\square$

$\bf{NOTE}$ This is a method that only closely estimates the value. It can be used in this question only because there are 180 terms and thus it would give a reasonably close estimate. Do not use this method if it has very little terms (like 3 or 4)

The idea is to find the average of all the terms from $sin(1°)$ to $sin(n°)$ First it must be noted that the maximum occurs when $n=180$ . So $\int _{ 0 }^{ \pi }{ sin(x)dx=2 }$ This is the area of the graph from $0$ to $180°$ Then the average can be found by assuming the area to be of a rectangle, with its base on the x-axis of length $\pi$ . Note this is only an estimate of the average of all of the $180$ terms. Therefore the total $\bf{estimated}$ value of the sum of all the terms is $\frac{2\times180}{\pi}=114.591559026$ So, rounding off to the nearest integer makes the answer $\boxed{115}$

Pretty close to the real answer: $114.5886501$

general formula is $\sum _{ k=0 }^{ n-1 }{ sin(a+kd) } =\frac { sin(\frac { nd }{ 2 } )sin(a+\frac { (n-1)d }{ 2 } ) }{ sin(\frac { d }{ 2 } ) }$ here we put a=d=1 and n=180 to obtain the sum to $sin({ 179 }^{ o })$ $(sin({ 180 }^{ o })$ can be excluded because it is equal to 0)

I used the python code

# Code by Shreyash
def f(n):
    res=0
    for i in xrange(1,n+1):
        res += math.sin(math.radians(i))
    return res 
math.ceil(max([f(n) for n in xrange(181)]))

Trigonometric identities are too hard! Let $\omega = e^{\left (i\cdot\frac{\pi}{180} \right )}=cis\left ( \frac{\pi}{180} \right )$ then we are interested in $\mathrm{Im}\left ( \omega ^{1}+\omega ^{2}+\ldots+\omega ^{180} \right )$ . Luckily, we can calculate the sum of geometric series, so: $\omega ^{1}+\omega ^{2}+\ldots+\omega ^{180}=\omega\left ( \frac{1-\omega^{180}}{1-\omega} \right )=\frac{2\omega}{1-\omega}$ and with enough will (or with Wolfram) you can calculate this number, and take its imaginary part.