Angle of elevation is 3.5. After a drive 13 miles coser, the angle of elevation is 9. The height of the tree?

Answer 1

I think that you are trying to find the height of a tree that is noticed to be tall enough at some distance to have an angle of elevation of 3.5; then when you move linearly towards said tree 13 miles, the elevation of the tree is 9.
Let the height of the tree be y
Let the initial distance from the tree (where you measured its elevation to be 3.5) be 13+x, since we do not know how far from the tree you are when you measure 9.
First eqn at elevation 3.5: tan(3.5) = y/(13+x)
Next eqn at elevation 9: tan(9) = y/x
Divide the first by the second:
tan(3.5)/tan(9) = x/(13+x) = constant, call it C, so that
x = (13+x)C = 13C + Cx
x(1-C) = 13C
x = 13C/(1-C) now we know x, so that by the second eqn above
y = x tan(9) = {13C/(1-C)}tan(9) the height of the tree; I leave it to you to evaluate C, tan(9), and y
y

Answer 2

Look like an awful tall tree. But you gave the number, so here goes. Let h be the height of the tree and x the distance from the first position to the tree. Then Tan 3.5= h/x.
At the 13 mile closer position, we have
Tan 9 = h/(x-13)
You can substitute equation 1 into equation 2 to get Tan 9 = h / [(h/Tan 3.5) -13], and
[ h Tan 9/Tan 3.5 - 13 Tan 9]=h, and
13 Tan 9 = h(Tan 9/Tan 3.5 -1)
13 Tan 9 /([Tan 9/Tan 3.5] -1)=h