Question from PRMO 2019. An ant leaves the anthill for its morning exercise. It walks 4 feet east and then makes a 160° turn 
to the right and walks 4 more feet. It then makes another 160° turn to the right and walks 4 more 
feet. If the ant continues this pattern until it reaches the anthill again, what is the distance in feet it would have travelled?
Well I have the solution but I am not able to understand it.
Let
$A_0 (0,0)$,$A_1 (4\cos0^\circ, 4\sin0^\circ)$, $A_2 (4\cos0^\circ + 4\cos160^\circ, 4\sin 0^\circ + 4\sin160^\circ)$,......, $A_n = (0,0)$ which gives 
$$4(\cos 0^\circ + \cos160^\circ +......+\cos(160(n-1))^\circ) = 0$$ and $$4(\sin 0^\circ +\sin 160^\circ +..... +\sin(160(n-1))^\circ) = 0$$ which in turn, gives
$$\sin((80n)^\circ) = 0$$
$$n = 9$$ Then distance covered =$4×9$ =$36\space \text{feet}$.
Please explain how do we get that $\cos 0$ and $\cos 60$ etc. in it.
This question has been asked in PRMO 2019 in India
 A: For instance, consider the variation in the $x$-coordinate of the position of the ant. It starts by walking $4$ units in a direction making an angle $0^\circ$ with the horizontal, so $\Delta x_1=4\cos0^\circ$, then walks $4$ units in a direction making an angle $160^\circ$ with the horizontal, and $\Delta x_2 =4\cos 160^\circ$. For the third turn, the direction would be along ($0^\circ +160^\circ+160^\circ$), and $\Delta x_3 = 4\cos(160\cdot 2)$. In general, you can deduce that the angle along which the ant walks on the $n^{th}$ turn would be the sum of the angles turned before, plus $160^\circ$. This gives $\Delta x_n = 4\cos\left(160(n-1)\right)$. We need $$\sum_{i=1}^n \Delta x_i =0 \implies \sum_{i=1}^n \cos\left(160(i-1)\right)=0 \implies n=9$$
Similarly for the $\Delta y_i$.
A: You need not go into applying the coordinate system and trigonometry here.
Much of PRMO is based on deduction and observations. By observation, you can see that the ant will form a star en route its path to the starting point with each internal angle of the star being $20^{\circ}$. Hence you can calculate the number of sides in the star to be $\frac{180^{\circ}}{20^\circ}=9$,
$\because\space$ The sum of all the internal acute angles of a regular star $=180^{\circ}$, which can be quite easily proven.
Hope this methodology clears your doubt!
