TRIGONOMETRY

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yuka

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Feb 16, 2006
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Jack and Jill are standing on the level ground 100 meters apart.

A large tree is due North of jack and on bearing of 063 from Jill.

The top of the tree appears at an angle of elevation of 25° to Jill and 15° to Jack.

Find the height of the tree.
 
Thank you!

but,, I can't draw this...

The top of the tree appears at an angle of elevation of 25° to Jill and 15° to Jack.

how to draw that??
 
This problem is a bit more complex than one might think ...

let the tree be at the origin of the xy plane

let Jack be a distance "b" due south of the tree

Jill will be in quadrant III since her bearing to the tree is 63 degrees. let her distance to the tree be "a". Using the bearing info, we can calculate the angle between a and b ... 90-63 = 27 degrees.

now, it is important to note that a < b because Jill has a higher angle of elevation to the top of the tree than Jack.

let the height of the tree be "h".

using right triangle trig with triangles in the plane vertical to the ground ...

tan(25) = h/a or h = a*tan(25)

tan(15) = h/b or h = b*tan(15)

so, now we know a*tan(25) = b*tan(15)

solving for one of the unknowns in terms of the other ...

a = b*tan(15)/tan(25) ... hang on to this equation for a second.

Now, using the triangle formed on the ground by Jack, Jill, and the tree, we can use the law of cosines ...

100<sup>2</sup> = a<sup>2</sup> + b<sup>2</sup> - 2ab*cos(27)

now, substitute b*tan(15)/tan(25) for a in the equation above ... and now you can solve for b.

Once you determine the value of b, you know that h = b*tan(15), and you can find the height of the tree.
 
Hello, yuka!

This is a three-dimensional problem, difficult to sketch in two dimensions.

\(\displaystyle A\) and \(\displaystyle B\) are standing on the level ground 100 meters apart.
A large tree is due north of \(\displaystyle A\) and on bearing of 063 from \(\displaystyle B\).
The top of the tree appears at an angle of elevation of \(\displaystyle 25^o\) to \(\displaystyle B\) and \(\displaystyle 15^o\) to \(\displaystyle A\).
Find the height of the tree.
Click to expand...
Looking down at the ground, the diagram looks like this:
Code: 
              T
      :       *
      :      /:
      :     / :
      :63° /  :
      :   /63°:
      :  /    :
      : /     :
      :/      :
      *       :
     B  \     :
          \   :
        100 \ :
              *
              A

My first 3-D sketch!
Code: 
                  S
                  o       + - - - - - - - - - - - - - - - - +
                  | * * /                                 /
                  |   *   *                             /
                 h| /   *     *                       /
                  |       *       *                 /
                / |         *         *           /
              /   o - - - - - * - - - - - o A   /
            /     T     *       *       /     /
          /                   *   *   / 100 /
        /                           o     /
      /                              B  /
    + - - - - - - - - - - - - - - - - +
\(\displaystyle h\,=\,ST = \text{height of tree}\)
\(\displaystyle AB\,=\,100\)
\(\displaystyle \angle ATB\,=\,63^o\)
\(\displaystyle \angle SAT\,=\,15^o\)
\(\displaystyle \angle SBT\,=\,25^o\)

I'll let someone else work out the solution . . . I need a nap!
 
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