SOLUTION: Two planes leave an airport at noon. One travels 800km/h S35degreesE and the other travels 750km/h S25degreesW. How far apart are they at 2:30pm?
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Question 1151498: Two planes leave an airport at noon. One travels 800km/h S35degreesE and the other travels 750km/h S25degreesW. How far apart are they at 2:30pm?
Thank you Answer by jim_thompson5910(35256) (Show Source):
Both planes take off at 12:00 pm noon, and they both take off from point A.
At 2:30 pm, plane #1 arrives at point B and plane #2 arrives at point C.
The flight duration so far is 2.5 hours.
Diagram
The notation means we start looking directly south and then turn 35 degrees toward the east as the diagram above shows for the blue angle.
The red angle is set up in a similar fashion.
For more information about compass bearings, see this page below. http://academic.brooklyn.cuny.edu/geology/leveson/core/linksa/comp.html
The red and blue angles combine to get 25+35 = 60 which is the green angle. This is angle A.
The sides b and c are adjacent to angle A.
side b is opposite angle B
side c is opposite angle C
The lowercase letters are used for side lengths; the uppercase letters for angles.
From the diagram:
A = 60
b = 1875
c = 2000
The values 1875 and 2000 are the distances each plane travels for their given speeds, and when the elapsed time is t = 2.5
Plane #1 travels at 800 km/hr for 2.5 hrs, so d = r*t = 800*2.5 = 2000
Plane #2 travels at 750 km/hr for 2.5 hrs, so d = r*t = 750*2.5 = 1875
The goal is to find the length of side 'a'. Use the law of cosines here
a^2 = b^2 + c^2 - 2*b*c*cos(A)
a^2 = 1875^2 + 2000^2 - 2*1875*2000*cos(60)
a^2 = 3515625 + 4000000 - 7500000*cos(60)
a^2 = 3515625 + 4000000 - 7500000*0.5
a^2 = 3515625 + 4000000 - 3750000
a^2 = 3765625
a = sqrt(3765625)
a = 1940.5218370325
a = 1940.52
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