Question

Two airplanes leave an airport at the same

Time. The velocity of the first airplane is
670 m/h at a heading of 43.7. The velocity
of the second is 550 m/h at a heading of 163 .
How far apart are they after 2.8 h?
Answer in units of m

Answer 1

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

Step-by-step 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 * 10^(6)\end{aligned}`.

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

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