We are given a triangle with one side $AB = 3.9$ km. We are given an exterior angle at vertex $F$ equal to $116^{\circ}$, and the length of the side $FB = 2.25$ km. We are given an exterior angle at vertex $B$ equal to $288^{\circ}$. We want to determine the length of the side $AF$.

1. Problem Description

We are given a triangle with one side $AB = 3.9$ km. We are given an exterior angle at vertex $F$ equal to $116^{\circ}$, and the length of the side $FB = 2.25$ km. We are given an exterior angle at vertex $B$ equal to $288^{\circ}$. We want to determine the length of the side $AF$.

2. Solution Steps

First, we need to find the interior angles at vertices $F$ and $B$.

The interior angle at $F$ is given by:

$angle(F) = 180^{\circ} - 116^{\circ} = 64^{\circ}$

The interior angle at $B$ is given by:

$angle(B) = 360^{\circ} - 288^{\circ} = 72^{\circ}$

Now we can find the angle at vertex $A$ using the fact that the sum of angles in a triangle is $180^{\circ}$:

$angle(A) = 180^{\circ} - angle(F) - angle(B) = 180^{\circ} - 64^{\circ} - 72^{\circ} = 180^{\circ} - 136^{\circ} = 44^{\circ}$

We can use the Law of Sines to find the length of the side $AF$:

$\frac{AF}{sin(B)} = \frac{AB}{sin(F)}$

$AF = \frac{AB * sin(B)}{sin(F)}$

$AF = \frac{3.9 * sin(72^{\circ})}{sin(64^{\circ})}$

$AF = \frac{3.9 * 0.9511}{0.8988}$

$AF = \frac{3.70929}{0.8988}$

$AF \approx 4.127$

3. Final Answer

$AF \approx 4.127$ km