SOLUTION: Use your knowledge of bearing, heading, and true course to sketch a diagram that will help you solve the problem. Two planes take off at the same time from an airport. The first

Algebra ->  Triangles -> SOLUTION: Use your knowledge of bearing, heading, and true course to sketch a diagram that will help you solve the problem. Two planes take off at the same time from an airport. The first      Log On


   

Question 1125963: Use your knowledge of bearing, heading, and true course to sketch a diagram that will help you solve the problem.
Two planes take off at the same time from an airport. The first plane is flying at 246 miles per hour on a course of 145.0°. The second plane is flying in the direction 165.0° at 357 miles per hour. Assuming there are no wind currents blowing, how far apart are they after 2 hours? (Round your answer to the nearest whole number.)
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
see my diagram.

$$$

the end result is triangle ABC.

angle A = 20 degrees.
AB = 2 * 246 = 492 miles.
AC = 2 * 357 = 714 miles.
the length of BC is the distance between the two planes after 2 hours.
that is found by using the law of cosines.
the law of cosines says [BC]^2 = [AB]^2 + [AC]^2 - 2 * [AB] * [AC] * cos(A).
apply that law and you get [AC] = 302.7449707 miles.

that's the distance between the planes after two hours.

angle A is equal to 165 degrees minus 145 degrees = 20 degrees.

i didn't really need the 90 degree angle, nor the 55 degree angle, nor the 75 degree angle, so you can disregard them.

in case you're wondering what i was doing, 145 = 90 + 55 and 165 = 90 + 75.
i was working my way down and then discovered i didn't need to do that.

the bearing of 145 degrees is the angle between the north vertical line and the line of travel of the first plane.

the bearing of 165 degrees is the angle between the north vertical line and the line of travel of the second plane.

the distances are the speed of the planes * 2 hours.