Two helicopters leave a heliport at the same time. The first flies 160km/h in a direction of 290 degrees. The second flies 100km/h in a direction of 250 degrees. After 3 hours, how far apart are the helicopters? Round your answer to the nearest kilometer.

315 km

Step-by-step explanation:

If the angle θ is measured from the x-axis, and we have a magnitude A, we can find the rectangular coordinates as:

Ax = A*cos(θ)

Ay = A*sin(θ)

Then:

The first helicopter has a speed of 160km/h, and it's angle is 290°.

Then the components will be:

Vx = 160km/h*cos(290°) = 54.7 km/h

Vy = 160km/h*sin(290°) = -150.4 km/h

Then the point location as a function of time, can be written as:

(where i assumed that the initial position of both helicopters was (0, 0))

P1(t) = (54.7 km/h*t, -150.4 km/h*t)

where t is time in hours.

We can do the same for the other helicopter:

The speed is 100km/h, and the angle is 250°

Then:

Vx = 100km/h*cos(250°) = -34.2 km/h

Vy = 100km/h*sin(250°) = -94 km/h

Then the point location as a function of time, will be:

P2(t) = (-34.2km/h*t, -94 km/h*t)

After 3 hours, the position of each helicopter will be:

P1(3h) = (54.7 km/h*3h, -150.4 km/h*3h) = (163.1 km, -451.2 km)

P2(3h) = (-34.2km/h*3h, -94 km/h*3h) = (-102.6km, -282km)

Now we can calculate the distance between these two points.

Remember that the distance between the points (a, b) and (c, d) is:

D = √( ( a - c)^2 + (b - d)^2)

In this case the distance will be:

D = √(  (163.1 km - (-102.6km))^2 + (-451.2 km - (-282km))^2)

D = √(  (163.1 km + 102.6km)^2 + (-451.2 km + 282km)^2)

D = 315km

i would say b) is your answer

sorry if im wrong

step-by-step explanation: