Two planes leave an airport at the same time. one airplane is flying 275 mph and the other is flying 375 mph. the angle between their flight paths is 55 degrees. after 3 hours, how far apart are they, to the nearest tenth show work

After 3 hours, both the airplanes are at a distance of 938.9 meters from each other.

Step-by-step explanation:

Speed of plane A = 275 mph

Speed of plane B = 375 mph

Angle between their flight path = Ф = 55°

Let their paths after 3 hours make a triangle.

Path of plane A is side 'a' = 275*3 = 825

Path of plane B is side 'b' = 375*3 = 1125

Distance between both planes after 3 hours = side 'c'

Here, we have two sides and one angle of the triangle.

We can find the third side using the law of cosines which is given as:

c² = a² + b² - 2ab*cosФ

c² = 825² + 1125² - 2(825)(1125)(cos(55°))

c² = 680625 + 1265625 - 1064745             where cos(55°) = 0.5736

c² = 881,505

c = √881,505

c = 938.9

c = 938.9 meters