SOLUTION: . An airplane is flying 10,500 ft. above the ground. The angle of depression from the plane to the base of a tree is 13o 50’. How far horizontally must the plane fly to be direct

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Question 1013227: . An airplane is flying 10,500 ft. above the ground. The angle of depression from the plane to the base of a tree
is 13o
50’. How far horizontally must the plane fly to be directly over the tree?

I've been trying to answer this question but I really don't understand it. I'm also not ale to attend the discussion. Can you help me?

Found 3 solutions by Alan3354, fractalier, MathTherapy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
An airplane is flying 10,500 ft. above the ground. The angle of depression from the plane to the base of a tree
is 13o
50’. How far horizontally must the plane fly to be directly over the tree?
=============
It's not clear what the angle is.

Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
Okay, sure...
Make yourself a diagram...the angle of depression from the plane is equal to the angle of elevation from the base of the tree...call it x...
We need to find the distance to the tree along the ground from under where the plane is...
We can use the tangent...50' of arc is .833 degrees
tan 13.833 degrees = 10500 / x
or
x = 10500 / tan 13.8333 = 42641 feet

Answer by MathTherapy(10556) About Me  (Show Source):
You can put this solution on YOUR website!
. An airplane is flying 10,500 ft. above the ground. The angle of depression from the plane to the base of a tree is 13o 50’. How far horizontally must
the plane fly to be directly over the tree?

I've been trying to answer this question but I really don't understand it. I'm also not ale to attend the discussion. Can you help me?
.
The horizontal distance (green broken line) the plane needs to travel to be directly above the tree
is the same as the horizontal distance (green solid line) directly below the plane to the base of the tree.
Let that distance be D
We then get: tan 13o 50’ = O%2FA
tan 13o 50’ = 10500%2FD
D %22%2A%22matrix%281%2C2%2Ctan+13%5Eo%2C+%2250%27%22%29+=+10500 ------- Cross-multiplying
D, or distance plane needs to travel = 10500%2Fmatrix%281%2C2%2Ctan+13%5Eo%2C+%2250%27%22%29%29 = highlight_green%2842641.2351%29 ft