I need help solving this word problem?

Question

Two planes take off at the same time from an airport. the first plane is flying at 296 miles per hour on a course of 160 degrees. The second plane is flying in the direction of 135 degrees at 249 miles per hour. Assuming there are no wind currents blowing, how far apart are they after 2 hours? Round your answer to the nearest mile.

Answer 1

You can use the Law of Cosines:

The first plane goes 592 miles and the second goes 498 miles during the 2 hours.

The angle between them is 160° - 135° = 25°

Let "x" be the distance between them (the length of the third side of the triangle).

Law of cosines: x² = 592² + 498² - 2(592)(498)(cos(25°))

x²= 64079.927 so x = √64079.927 = 253 miles

Answer 2

HI Olivia,

Two planes take off at the same time from an airport. the first plane is flying at 296 miles per hour on a course of 160 degrees. The second plane is flying in the direction of 135 degrees at 249 miles per hour. Assuming there are no wind currents blowing, how far apart are they after 2 hours? Round your answer to the nearest mile.

For word problems, gather the information you are given

Plane 1) 296mph at 160 degrees

Plane 2) 249mph at 135 degrees

The leave at same time from same location. There are no winds affecting their flight.

Find: How far apart are they after 2 hours

Strategy:

Draw a picture of the planes as vectors.

Each plane will fly for 2 hours. Find out where each plane is at 2 hours and break down the location into the x and y coordinate.

Use the distance formula between them to determine how far apart they are.

Plane 1 will travel 2 hours at 296mph This means it will travel 2 hrs X 296mph = 592 miles

The angle is 160 degrees. The reference angle is (180-160) = 20 degrees

Plane 2 will travel 2 hours at 249mph. This means it will travel 2 hrs X 249mph = 498 miles

The angle is 135 degrees. The reference angle is 180-135 = 45 degrees

(here is a pic from geogebra (geogebra.org/calculator/tm3n4ktp)

Now you need to break down each vector (592 at 20 degrees and 498 at 45 degrees) into their x and y values

y = h sinθ y = y distance , h = hypotenuse in this case its the distance the plane traveled.

x = h cosθ x=x distance

Plane 1

y1 = 592 sin 20 = 202.48 miles

x1 = 592 cos 20 = -556.30 miles (it's in quadrant 2 so the x value is negative)

SO they are at (-556.30, 202.48)

Plane 2

y2 = 498 sin 45 = 352.14

x2 = 498 cos 45 = -352.14 (it's in quadrant 2 so the x value is negative)

(-352.14, 352.14)

Use the distance formula between the two locations

distance = sqrt [(x2-x2)^2 + (y2-y1)^2]

d = sqrt [(-352.14 - (-556.30))^2 + (352.14-202.48)^2]

d= sqrt (41681.3056 + 22398.1156)

d= 253.1391341 miles

Rounded to the nearest mile = 253 miles

Hope this helps. Let me know if you have any more questions!!