From a plane flying due east at 265m above sea level, the angles of depression of two ships sailing due east measure 35* and 25*. how far apart are the ships?
Let's assume the distance between the plane and the first ship is x, and the distance between the plane and the second ship is y. Since the plane is flying due east, it is at the same level as the first ship.
We can create two right-angled triangles to represent the situation. In the first triangle, the angle of depression is 35°, and the height of the plane (which is the same as the height of the first ship) is 265m. In the second triangle, the angle of depression is 25°, and the height of the plane (which is the same as the height of the second ship) is 265m.
In the first triangle:
tan 35° = 265/x
x = 265/tan 35°
In the second triangle:
tan 25° = 265/y
y = 265/tan 25°
Therefore, the distance between the ships is x + y:
Distance = (265/tan 35°) + (265/tan 25°)