In this question, the path is a regular hexagon $A B C D E F$ of side length $500 \mathrm{~m}$. In Fig,


Let the motorist start from $A$.
Third turn
The motorcyclist will take the $3^{\text {rd }}$ turn at $D$. Displacement vector at $D=A D$
Magnitude of this displacement $=500+500$ $=1000 \mathrm{~m}$
Total path length from $A$ to $D=A B+B C+C D$ $=500+500+500=1500 \mathrm{~m}$
Sixth turn
The motorcyclist will take the $6^{\text {th }}$ turn at $A$.
$\therefore \quad$ Displacement vector is null vector.
Magnitude of this displacement $=500+500$ $=1000 \mathrm{~m}$
Total path length $=A B+B C+C D+D E+E F$ $=500+500+500+500+500+500=3000 \mathrm{~m}$
Eighth turn
The motorcyclist takes the $8^{\text {th }}$ turn at $C$.
$\therefore \quad$ Displacement vector $=A C$, which is represented by the diagonal of the parallelogram $A B C G$.
$$
\therefore \sqrt{\left[(500)^2+(500)^2+2 \times(500) \times(500) \cos 60^{\circ}\right]}
$$
$$
=\sqrt{\left[(500)^2+(500)^2+250000\right]}=866.03 \mathrm{~m}
$$
$\tan \beta=500 \sin 60^{\circ} \%\left\{500+500 \cos 60^{\circ}\right]$
$$
\begin{aligned}
&=(500 \sqrt{3} / 2) /\{500(1+1 / 2)\}=1 / \sqrt{3} \\
&=\tan 30^{\circ} \text { or, } \beta=30^{\circ}
\end{aligned}
$$