Posted by Doni on Saturday, May 17, 2008 at 12:40pm.
This is a vector problem. Let the x axis be the direction that the 300 km/h plane flies. Assume the faster plane flies at 75 degrees to that direction with a velocity component along both the +x and +y axes.
The location of plane 1 at time t is given by
X1 = 300 t
Y1 = 0
The location of plane 2 at time 2 is
X2 = 420 cos 75 t = 108.7 t
Y2 = 420 sin 75 t = 405.7 t
The distance between than at any time t is
sqrt [(X2-X1)^2 + (Y2-Y1)^2]
Plug in t = 3 hours and solve
I n my previous answer, I solved it using a vector method, not a trig method. Both will give the same answer. To use trigonometry, draw a triangle with two sides representing the two distances and directions of the planes after three hours. The angle between these two sides is 75 degrees. The third side of the triangle is the line between the planes at that time.
You have two sides of a triangle and the included angle (75 degrees). The law of cosines can be used for the third side, which is what you want.
please show me again in trig way
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