Question

NEED HELP DUE AT 11:59

Two airplanes leave an airport at the same

time. The velocity of the first airplane is

740 m/h at a heading of 39.3

◦

. The velocity

of the second is 620 m/h at a heading of 87◦

.

How far apart are they after 3.5 h?

Answer in units of m.

Answer 1

1962.6 m (nearest tenth)

Step-by-step explanation:

`\boxed{\sf Distance=Velocity * Time}`

First airplane

• Velocity = 740 m/h
• Time = 3.5 h
• Distance = 740 × 3.5 = 2590 m
• Angle = 39.3°

Second airplane

• Velocity = 620 m/h
• Time = 3.5 h
• Distance = 620 × 3.5 = 2170 m
• Angle = 87°

Draw a diagram with the calculated distances and given angles (see attachment).

To find how far apart the planes are, connect the ends of the line segments to create a triangle and use the cosine rule to find x.

Cosine rule

`\boxed{c^2=a^2+b^2-2ab \cos C}`

`\textsf{Where $a, b$ and $c$ are the sides and $C$ is the angle opposite side $c$}.`

From inspection of the diagram (attached):

• a = 2170
• b = 2590
• c = x
• C = 87° - 39.3° = 47.7°

Substitute the values into the formula and solve for x:

`\begin{aligned}\implies x^2 & =2170^2+2590^2-2(2170)(2590) \cos 47.7^(\circ)\\x^2 & =4708900+6708100-11240600\cos 47.7^(\circ)\\x^2 & =11417000-11240600\cos 47.7^(\circ)\\x^2 & =3851935.541...\\x & =√(3851935.541...)\\x & = 1962.634846...\end{aligned}`

Therefore, the airplanes are 1962.6 m apart after 3.5 hours (nearest tenth).