Q:

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 tenthPlease show work

Accepted Solution

A:
Answer: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 mphSpeed of plane B = 375 mphAngle between their flight path = Ф = 55°Let their paths after 3 hours make a triangle.Path of plane A is side 'a' = 275*3 = 825Path of plane B is side 'b' = 375*3 = 1125Distance 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*cosФc² = 825² + 1125² - 2(825)(1125)(cos(55°))c² = 680625 + 1265625 - 1064745             where cos(55°) = 0.5736c² = 881,505c = √881,505c = 938.9c = 938.9 meters