Answer:
Approximately [tex]1055\; {\rm m}[/tex].
Explanation:
Let [tex]{\rm AB}[/tex], [tex]{\rm BC}[/tex], and [tex]{\rm AC}[/tex] denote the length of the sides of triangle [tex]\triangle {\rm ABC}[/tex]. Let [tex]\angle {\rm A}[/tex] denote the measure of angle [tex]{\rm A}[/tex]. By the Law of Cosines:
[tex]({\rm BC})^{2} = ({\rm AB})^{2} + ({\rm AC})^{2} - 2\, ({\rm AB})\, ({\rm AC})\, \cos(\angle A})[/tex].
Take the square root of both sides to find the length of segment [tex]{\rm BC}[/tex].
In this question, let [tex]{\rm A}[/tex] denote the position of the airport. Let [tex]{\rm B}[/tex] and [tex]{\rm C}[/tex] denote the position of the aircrafts after [tex]2.8\; {\rm h}[/tex]. Join [tex]{\rm A}\![/tex], [tex]{\rm B}\![/tex], and [tex]{\rm C}\![/tex] to obtain a triangle (refer to the attached diagram.)
The length of segment [tex]{\rm AB}[/tex] would be [tex](2.8)\, (670) = 1876[/tex].
The length of segment [tex]{\rm AC}[/tex] would be [tex](2.8)\, (550) = 1540[/tex].
The measure of angle [tex]{\rm A}[/tex] would be [tex]\angle {\rm A} = (163 - 43.7)^{\circ} = 119.3^{\circ}[/tex].
Find the length of segment [tex]{\rm BC}[/tex] with the Law of Cosines:
[tex]\begin{aligned}({\rm BC})^{2} &= ({\rm AB})^{2} + ({\rm AC})^{2} - 2\, ({\rm AB})\, ({\rm AC})\, \cos(\angle A}) \\ &= 1876^{2} + 1540^{2} - 2\, (1876)\, (1540)\, \cos(119.3^{\circ}) \\ &\approx8.71867 \times 10^{6}\end{aligned}[/tex].
[tex]\begin{aligned}({\rm BC}) \approx \sqrt{8.71867 \times 10^{6}} \approx 2953\end{aligned}[/tex].
Therefore, the distance between the aircrafts would be approximately [tex]2953\; {\rm m}[/tex] after [tex]2.8\; {\rm h}[/tex].
