SOLUTION: Two planes have left the airport. One plane is 30 miles wesst of the airport and has an altitude of 4 miles. The other is 30 miles east of the aiport and has altitude of 2 miles. W
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Question 734054: Two planes have left the airport. One plane is 30 miles wesst of the airport and has an altitude of 4 miles. The other is 30 miles east of the aiport and has altitude of 2 miles. What is the distance between the planes? Found 2 solutions by ankor@dixie-net.com, Alan3354:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Two planes have left the airport.
One plane is 30 miles west of the airport and has an altitude of 4 miles.
The other is 30 miles east of the airport and has altitude of 2 miles.
What is the distance between the planes?
:
find the angles to the planes using the right triangles formed with the ground
Let A = angle of the higher plane
Let B = angle of the lower
;
Sin(A) =
A = 7.66 degrees
Sin(B) =
B = 3.82 degrees
:
Then the angle opposite the distance between the isosceles triangle formed by the base and the two aircraft: 180 - 7.66 - 3.82 = 168.5 degrees
the other two angles of this triangle: = 5.75 degrees
:
Use the law of sines to find the distance (d) between the aircraft =
Cross multiply and find the sines
.1d = 30 * .1994
d =
d = 59.81 mi between the aircraft
You can put this solution on YOUR website! Two planes have left the airport. One plane is 30 miles wesst of the airport and has an altitude of 4 miles. The other is 30 miles east of the aiport and has altitude of 2 miles. What is the distance between the planes?
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The distance on the ground = 10+30 = 60 miles
The vertical distance = 2 miles
d =~ 60.00333 miles
