Question 78320
<pre><font size = 5><b> 
From a point A on the ground, the angle of 
elevation to the top of a tall building is 
24.1 degrees. From a point B, which is 600 
ft closer to the building, the angle of 
elevation is measured to be 30.2 degrees. 
Find the height of the building.
                        
                       *|C
                 * *    |
           *            |h
    *      *            | 
A_____B_________________|D
  600         x

I can't draw slanted lines on here, but you can
on your paper.  I've tried to indicate AC and BC
with asterisks " * " above.

Let C be the top of the building and D the bottom of
the building.

There are two right triangles, CAD and CBD.
Let h = CD and x = BD. We are given that AB = 600.

So from right triangle CAD we have:

                 h
tan(24.1°) = ----------
              600 + x


and from right triangle CBD we have:

              h
tan(30.2°) = ---
              x

To make the algebra easier to write, 
let's let:
 
T = tan(24.1°) and U = tan(30.2°)
Then those two equations are

               h
       T = --------- 
            600 + x

and

            h
       U = --- which means h = Ux
            x

Clearing of fractions in those equations
gives these two equations:

       h = T(600 + x)
       h = Ux

Since both right sides equal h, we set them
equal to each other

      Ux = T(600 + x)

Now we solve for x:

      Ux = 600T + Tx

 Ux - Tx = 600T

x(U - T) = 600T

Divide both sides by (U - T)

             600T
       x = ---------
            U  -  T 

Now we now go back and substitute
for U and T

T = tan(24.1°) and U = tan(30.2°)

and we have:

                  600·tan(24.1°)
       x = ---------------------------
            tan(30.2°)  -  tan(24.1°)


Punch that out on your calculator and you get

       x = 1992.638095

Then we can find h from

       h = Ux
       h = tan(30.2°)(1992.638095)
       h = 1159.743065

I'd round that off to 1160 feet.

Edwin</pre>