143. What is the sum of the exterior angles at the five vertices of a convex pentagon?
$\checkmark$ Together with the 5 interior angles (which sum up to $540^{\circ}$), we get 5 straight angles, that is, $900^{\circ}$, so the sum of the exterior angles is $900-540=360^{\circ} . \triangleleft$

Similarly, one can verify that the sum of the exterior angles of a convex hexagon, heptagon, and generally any $n$-sided polygon is $360^{\circ}$. Indeed, together with the interior angles, we get $n$ straight angles, that is, $n \cdot 180^{\circ}$, and the interior angles sum up to $(n-2) \cdot 180^{\circ}$, so the difference, which is the sum of the exterior angles, is $2 \cdot 180=360^{\circ}$.

However, there is another, more intuitive (though less formal) explanation. Imagine a car that drives around the perimeter of the polygon. The angles of the polygon are the points where the car turns. At each point, it turns by an angle equal to the exterior angle of the polygon.
![](https://cdn.mathpix.com/cropped/2024_05_21_90703b5d5e76e3b5cd3dg-052.jpg?height=560&width=284&top_left_y=454&top_left_x=1434)
After completing the circuit, the car will have made a full rotation, that is, it will have turned $360^{\circ}$.
![](https://cdn.mathpix.com/cropped/2024_05_21_90703b5d5e76e3b5cd3dg-053.jpg?height=220&width=306&top_left_y=507&top_left_x=338)