At 1 pm, an airplane left an airport at a speed of 800 kph and is heading N 35 degrees E. An hour later, anothe airplane left at a speed of 1000 kph heading directly East. How far apart are the planes at 4 pm?
The first airplane leaves at 1 pm at a speed of 800 kmph traveling North 35 degrees East. At 2 pm another airplane leaves at 1000 kmph traveling directly East.
At 4 pm the distance traveled by the first plane towards the East is equal to 800*3*sin 35 = 1376.58 km. The distance traveled towards the North is 800*3*cos 35 = 1965.96 km. The distance traveled towards the East by the second airplane is 1000*2 = 2000 km.
The distance between the two airplanes at 4 pm is equal to `sqrt( 1965.96^2 + (2000 - 1376.58)^2)` = 2062.43 km
The two airplanes are 2062.43 km away from each other at 4 pm.