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?

*[illustration Travel_No_1013229].
<pre>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 13<sup>o</sup> 50’ = {{{O/A}}}
tan 13<sup>o</sup> 50’ = {{{10500/D}}}
D {{{"*"}}}{{{matrix(1,2,tan 13^o, "50'") = 10500}}} ------- Cross-multiplying
D, or distance plane needs to travel = {{{10500/matrix(1,2,tan 13^o, "50'"))}}} = {{{highlight_green(42641.2351)}}} ft