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 1013229: . 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 Theo, macston, MathTherapy:
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
i'm not sure where the 50 feet comes from.
let me see what i can do without that.

the plane is flying at 10,500 feet.

the angle of depression from the plane to the base of the tree is 13 degrees, if i understand that correctly.


the angle of depression is the angle that is formed by drawing a line that is 13 degrees from the horizontal line formed by the plane.

see the following reference for a discussion of angle of elevation and angle of depression.

http://www.purplemath.com/modules/incldecl.htm

you have two horizontal lines at AD and CB.
you have two vertical lines at AC and DB.

the length of line AC is equal to the length of DB.
the length of AD is equal to the length of CB.

the angles at A, B, C, D, are all 90 degrees.

ADBC is therefore a rectangle.

the length of AD and CB is represented by x in the diagram.

since line AD and CB are parallel, and since line AB is a transversal of those two parallel lines, then angle DAB is congruent to angle ABC, so they're both equal to 13 degrees.

tan(angle ABC) = tan(13) = 10500 / x.

solve for x to get x = 10500 / tan(13) = 45480.49668.

x is the horizontal distance from the plane to the point D directly above the foot of the tree at point B.

that's your solution.

x = 45480.49668 feet.

round as you see fit.

here's the diagram of the triangles and rectangle formed.

$$$



Answer by macston(5194)   (Show Source): You can put this solution on YOUR website!
.
Tan (13 degrees 50 minutes) = altitude / horizontal distance
horizontal distance = altitude/tan 13.833 degrees
horizontal distance = 10500ft/tan 13.833 degrees
horizontal distance = 42641 ft
.


Answer by MathTherapy(10556)   (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’ = 

tan 13o 50’ = 

D  ------- Cross-multiplying

D, or distance plane needs to travel =  =  ft 

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