Question

Two airplanes leave an airport at the same

time. The velocity of the first airplane is
730 m/h at a heading of 56.8◦
. The velocity
of the second is 550 m/h at a heading of 111◦
.
How far apart are they after 2.9 h?
Answer in units of m.

Answer 1

1753.67 m

Step-by-step explanation:

Using the law of cosine

`c=\sqrt {(a^(2)+ b^(2)-2abcos\theta_(c))}`

After 2.9 hours, the first plane has moved 730*2.9=2117 m, let this be a

The second plane has moved 550*2.9=1595 m, let this be b

The angle between them is 111-56.8=54.2, let this be
`\theta_(c)`

`c=\sqrt {(2117^(2)+ 1595^(2)-2abcos 54.2)}`

c=1753.67 m