Thread: The Law of Cosines

I am having problems with this problem. Any help would be greatly appreciated.

Two airplanes leave an airport at the same time. The first flies 150 km/h in a direction of 320 degrees. The second flies 200 km/h in a direction of 200 degrees. After 3 hr, how far apart are the planes?

I am having a problem with the 200 degrees flight also, but, ....

Imagine, or draw the figure on paper.
Flight directions are measured from positive y-axis going clockwise.

If we call the quadrants in the usual way,
320 degrees is in the 2nd quadrant,
200 degrees is in the 3rd quadrant.

The angle from 200deg to 320deg is
320 -200 = 120 degrees.

distance = rate * time
So, after 3 hrs,
150 km/h *3h = 450 km
200 km/h *3h = 600 km

Hence, the figure is an obtuse triangle ABC where
side AB = 450 km
angle ABC = 120 degrees
side BC = 600 km
side CA = distance between the two planes

That is two known sides and their included angle.

By Law of Cosines,
(CA)^2 = (AB)^2 +(BC)^2 -2(AB)(BC)cos(120deg)
(CA)^2 = (450)^2 +(600)^2 -2(450)(600)cos(120deg)
(CA)^2 = 202,500 +360,000 -540,000(-0.5)
(CA)^2 = 202,500 +360,000 +270,000
(CA)^2 = 832,500
CA = sqrt(832,500)
CA = 912.4

Therefore, after 3 hours, the two planes are 912.4 km apart. ---answer.