the pilot of a plane flying 2000 meters above sea level observes two ships in line due east of the plane. the angles of depression of the line of sight of the pilot to the two ships are 42 degrees and 30 degrees. how far apart are the two ships at the time of the sighting?
To find the distance between the two ships, we can use trigonometry and the concept of angles of depression.
Let's denote the distance between the pilot and one of the ships as x, and the distance between the pilot and the other ship as y. We can now form two right-angled triangles with the pilot at the vertex, one for each ship.
In the first triangle, the angle of depression is 42 degrees, while in the second triangle, the angle of depression is 30 degrees. The height of the pilot above sea level is given as 2000 meters.
Using trigonometry, we know that the tangent of an angle is equal to the opposite side divided by the adjacent side. In this case, the opposite side is the height of the pilot above sea level (2000 meters).
For the first triangle (with an angle of depression of 42 degrees):
tan(42 degrees) = 2000 / x
For the second triangle (with an angle of depression of 30 degrees):
tan(30 degrees) = 2000 / y
To find x and y, we need to isolate them in each equation:
x = 2000 / tan(42 degrees)
y = 2000 / tan(30 degrees)
Now, we can calculate the values of x and y:
x ≈ 2000 / 0.9004 ≈ 2,220.4 meters
y ≈ 2000 / 0.5774 ≈ 3,464.1 meters
Therefore, the two ships are approximately 2,220.4 meters apart at the time of the sighting.