Find the distance between the two ships

A plane is flying at an elevation of 35000 ft. The angles of depression to the two ships are 40 degrees and 52 degrees.

Draw the picture.

35000 ft is the opposite of each angle. The horizontal distance to each ship is the adjacent of that angle.

So tan(40) = 35000/d1
tan(52) = 35000/d2

Calculate the two distances.

....../\
..../....\....Height=3500
../.......\
S1.......S2
a1=40.a2=52

Dist between ships = x
Dist from S1 to plane (horizontally) = y

OppS1 = 3500
OppS2 = 3500
AdjS1 = y
AdjS2 = x-y

Cot = Adj/Opp

Cot(40deg) = y/3500
Cot(52deg) = (x-y)/3500

y = 3500Cot(40deg)
Cot(52deg) = (x-3500Cot(40deg))/3500

Distance between ships = x = 3500[Cot(52deg)+Cot(40deg)] ≈ 6.905637267*10^3ft

tanΘ1 = A/d1
tanΘ2 = A/d2

d1 = A/tanΘ1
d2 = A/tanΘ2
d1-d2 = A(1/tanΘ1 - 1/tanΘ2) = A(cotΘ1 - cotΘ2)

d1-d2 = 35000(cot40 - cot52) ≈ 14366 ft ≈ 4789 yards (the normal nautical distace specification) ≈ 2.72 statute miles ≈ 2.36 nautical miles

well you have one right triangle with one leg 35000 and degrees of 50, 90, 40
another right triangle with one leg 35000 and degrees of 38, 90, 52

( 35000 / sin 40 = x / sin 50 )
solve for x, thats the distance.

Distance = 35000 * Tan 52 - 35000* Tan40 = 35000(tan52 - tan 40) = 15429 feet (to nearest ft)