Approximating the angle between the trajectory I started to learn physics this semester and I found the following task:
A contestant is participating in a half-maraton tournament(straight line length $L =21095$ meter) running in a zig-zag manner (constantly surpassing other contestants), holding a stable angle $\alpha$ between the trajectory. After finishing the run, the contestant has noticed that the distance travelled was 500 meters longer than $L$. Approximate the angle $\alpha$ without using a calculator.
I have no idea how to approach this problem so any help would be useful.
 A: 
The green line is the line that the "zig-zag" runner is running.
Thus:
$$s\cos(\alpha)=a\tag 1$$
So if he is doing this n times during the half-marathon you obtain
that:
$$n\,(s-a)=\Delta L\tag 2$$
With Eq. (1) and (2) you obtain that:
$$\cos(\alpha)=\frac{a\,n}{\Delta L+a\,n}$$
Edit
with the remarks from @AgniusVasiliauskas
$$a\,n=L~\Rightarrow \\(\alpha)=\frac{L}{\Delta L+L}\approx 0^\circ $$
A: The angle is small but it is not zero. The problem relies on the fact that the angle is small when it asks to find an approximate value without using a calculator.
You find $cos(\alpha)=\frac{L}{L+\Delta L} \approx 0.977 $.
But, for small angles, you can approximate $cos(\alpha) \approx 1-\frac{\alpha^2}{2} $
So, by comparing the two relationships, you can get the angle in radians as
$\alpha \approx 0.21 rad$.
To find it in degrees you just multiply by $180/\pi$ (or about 57 or even more roughly, 60) and indeed you get an angle of aproximately $12^0$.
If you are not familiar with the Taylor expansion for cosine function, you can just use $sin^2 +cos^2 =1 $  to find $sin(\alpha)$ and then use the fact that for small angles $sin(\alpha) \approx \alpha$. This is also a result of a Taylor expansion hovewer it is used in introductory Physics in the study of the pendulum so it is more familiar for the students, I believe.
