Question 250653
if I understand this correctly, the scenario is as follows:


<pre>

           A   x  <--- top of tree
               x   x
               x      x
               x         x
               x            x
               x               x     D
               x                  x  
               x                     x   <--- top of joe's head
               x                     x   x
               x                     x      x
               x          1.59m -->  x         x
               x                     x            x
               x                     x               x
               x                     x                  x
           B   x   x   x   x   x   x   x   x   x   x   x   x   C
               |--------23.56m-------|--------8.42m--------|
                                     
                                     E


</pre>
If this is the case, then we can first find the angle DCE and then find the height of the tree.


Tan (DCE) = opposite / adjacent = 1.59 / 8.42 = .188836105


Angle DCE = Arctan (.188836105) = 10.69359057 degrees.


Tan (ACB) = Tan (10.69359057) = .188836105


Tan (ACB) = opposite / adjacent = AB / BC = AB / (23.56 + 8.42) = AB / 31.98


Tan (ACB) = AB / 31.98


Since Tan (ACB) is the same as Tan (DCE), this formula becomes:


.188836105 = AB / 31.98


multiply both sides of this equation by 31.98 to get:


31.98 * .188836105 = AB


AB = 6.038978622 meters high.


Since triangle DCE is similar to triangle ACB, then the corresponding sides are proportional to each other so you could have used a ratio to find the same answer.


The ratio would have been:


DE / CE = AB / CB


This would have come out as:


1.59 / 8.42 = x / 31.98


Cross multiply to get 8.42*x = 1.59 * 31.98


divide both sides by 8.42 to get x = (1.59 * 31.98) / 8.42 = 6.038978622.