Question

Two airplanes leave an airport at the same

Answer 1

Answer:

1753.67 m

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