SOLUTION: From a plane flying due East at 265 meters above sea level, the angles of depression of two ships sailing due East measure 35� and 25�. How far apart are the ships?

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Question 1035787: From a plane flying due East at 265 meters above sea level, the angles of depression of two ships sailing due East measure 35� and 25�. How far apart are the ships?
Found 2 solutions by Boreal, MathTherapy:
Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website!
This has to be drawn.
There is a line from the plane to the distant ship with depression 25 degrees. Call this line c. We know the distance to that ship for a right triangle formed by the ship to the plane's height (265 m) the course of the plane and c. The sin 25=265/c, and c=265/sin (25), or 627.04 m.
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Similarly, for the closer ship, there is line a, its distance from the plane, which is 265/sin(35)=462.01 m. This is line a.
Let b be the line between the ships. Angle A is the angle between the ship and the plane, which is 25 degrees (alternate interior angles). Angle B, opposite to the inter-ship distance, is 10 degrees, the difference in depression angles. Angle C is their sum subtracted from 180 degrees, or 155 degrees.
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Use law of cosines for side b
b^2=a^2+c^2-2ac cos B
b^2=462.01^2+627.4^2-2(462.01)(627.04)cos(25)
=606632.4-525112.5=81,519.9
b=285.5 m

Answer by MathTherapy(10557) (Show Source):
You can put this solution on YOUR website!

From a plane flying due East at 265 meters above sea level, the angles of depression of two ships sailing due East measure 35� and 25�. How far apart are the ships?

Distance from the farther ship () to the point DIRECTLY below the plane:
Distance from the closer ship () to the point DIRECTLY below the plane:
Distance between the ships: . This is so, assuming the ships are on the same side of the plane