Question

Two airplanes leave an airport at the same

Answer 1

Answer:

Approximately $1055\; {\rm m}$.

Explanation:

Let ${\rm AB}$, ${\rm BC}$, and ${\rm AC}$ denote the length of the sides of triangle $\triangle {\rm ABC}$. Let $\angle {\rm A}$ denote the measure of angle ${\rm A}$. By the Law of Cosines:

$({\rm BC})^{2} = ({\rm AB})^{2} + ({\rm AC})^{2} - 2\, ({\rm AB})\, ({\rm AC})\, \cos(\angle A})$.

Take the square root of both sides to find the length of segment ${\rm BC}$.

In this question, let ${\rm A}$ denote the position of the airport. Let ${\rm B}$ and ${\rm C}$ denote the position of the aircrafts after $2.8\; {\rm h}$. Join ${\rm A}\!$, ${\rm B}\!$, and ${\rm C}\!$ to obtain a triangle (refer to the attached diagram.)

The length of segment ${\rm AB}$ would be $(2.8)\, (670) = 1876$.

The length of segment ${\rm AC}$ would be $(2.8)\, (550) = 1540$.

The measure of angle ${\rm A}$ would be $\angle {\rm A} = (163 - 43.7)^{\circ} = 119.3^{\circ}$.

Find the length of segment ${\rm BC}$ with the Law of Cosines:

$\begin{aligned}({\rm BC})^{2} &= ({\rm AB})^{2} + ({\rm AC})^{2} - 2\, ({\rm AB})\, ({\rm AC})\, \cos(\angle A}) \\ &= 1876^{2} + 1540^{2} - 2\, (1876)\, (1540)\, \cos(119.3^{\circ}) \\ &\approx8.71867 \times 10^{6}\end{aligned}$.

$\begin{aligned}({\rm BC}) \approx \sqrt{8.71867 \times 10^{6}} \approx 2953\end{aligned}$.

Therefore, the distance between the aircrafts would be approximately $2953\; {\rm m}$ after $2.8\; {\rm h}$.