Question 734054
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) = {{{4/30}}}
A = 7.66 degrees
Sin(B) = {{{2/30}}}
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: {{{(180-168.5)/2}}} = 5.75 degrees
:
Use the law of sines to find the distance (d) between the aircraft
{{{d/sin(168.5)}}} = {{{30/sin(5.75)}}}
Cross multiply and find the sines
.1d = 30 * .1994
d = {{{5.981/.1}}}
d = 59.81 mi between the aircraft