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
Two planes leave an airport at the same time, one flying at 300 km/h and the other at 420 km/h The angle between their flight paths is 75 degrees. Aftar three hours, how far apart are they ?
i guss i have to use the equation D= b.h but i cant fit the number in spot can u please help me
3 answers
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.
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
1 answer