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contestada

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.

Respuesta :

opudodennis

Answer:

1753.67 m

Explanation:

Using the law of cosine

[tex]c=\sqrt {(a^{2}+ b^{2}-2abcos\theta_{c})}[/tex]

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 [tex]\theta_{c}[/tex]

[tex]c=\sqrt {(2117^{2}+ 1595^{2}-2abcos 54.2)}[/tex]

c=1753.67 m