Two airplanes leave an airport at the same time, the first headed due north and the second at a bearing of N42^ E . At 2:00PM, the first airplane is 312 miles from the airport while the second airplane is 487 miles from the airport. Assuming both followed linear paths from the airport, how far apart are the two airplanes at 2:00PM?

The two airplanes are about 330miles apart.

Step-by-step explanation:

The diagram interpreting the question has been attached to this response.

As shown in the diagram,

i. the airplanes leave at point C.

ii. at 2.00pm the first and second airplanes are at points A and B respectively, where they are 312miles and 487miles away from the starting point C in directions due north and N42E from the point C.

iii. the points A, B and C form a triangle with sides a, b and c.

To solve for the value of c which is the distance between the two planes at 2.00pm, the cosine rule is used.

c² = a² + b² - 2abcosC            --------------(i)

where;

b  = 312miles

a = 487miles

C = 42°

Substitute these values into equation (i) and solve as follows;

c² = (487)² + (312)² - 2(312)(487)cos(42)

c² = (237169) + (97344) - 303888cos(42)

c² = (237169) + (97344) - 303888(0.7431)

c² = 334513 - 225819.1728

c² = 108693.8272

Take the square root of both sides

√c² = √108693.8272

c = 329.69

c ≅ 330miles

Therefore, the two airplanes are far apart by 330miles

< 4, -5>

step-by-step explanation:

starting from g. we move 4 right and then 5 down to get to h. that means the vector is < 4, -5>

i can you with this question sweetie!

step-by-step explanation: