Question

Two airplanes leave an airport at the same time, one going NW bearing 135 degrees at 415mph and the other going east at 327 mph. How far apart are the planes after 4 hours (to the nearest mile)?

Answer 1

Answer:

Step-by-step explanation:

in 4 hrs

NW plane moves=415*4=1660 miles

East plane moves=327*4=1308 miles

both directions makes angle=90+45=135° with each other.

so we have to find the third side of a triangle opposite 135°

$cos~135=\frac{1660^2+1308^2-x^2}{2 \times 1660 \times 1308} \\-\frac{\sqrt{2} }{2} \times 2\times 1660 \times 1308=2755600+1710864-x^2\\ x^2=4466464+2171280\sqrt{2} \approx 4466464+3070654 \approx 7537118\\x=\sqrt{7537118} \approx 2745 miles$

Answer 2

The distance between the two planes after four hours is 2745.381m

Data;

• angle = 135 degree
• v1 = 415mph
• v2 = 32mph
• time = 4 hours

Cosine Rule

To find the distance between the after 4 hours,

Let AB = 415mph, in 4 hours; = 4 * 415 = 1660m = a

Let BC = 327mph, in 4 hours; = 4 * 327 = 1308m = b

We need to find the distance between the two plane = x

$x^2 = 1660^2 + 1308^2 - 2(1660)(1308)cos135\\x^2 = 7537117.6237\\x = \sqrt{ 7537117.6237}\\x = 2745.381$

The distance between the two planes after four hours is 2745.381m

Learn more on cosine rule here;

https://brainly.com/question/4372174