Question 1014220
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There are two towers. One at A and one at B. A and B are are two points on the same horizontal ground and 120 m apart. 
The angle of elevation of the top of the tower at B as observed from A is twice that of the tower at A as observed at B. 
At C, midway between A and B, the angles of elevation are complimentary. Find the heights of the the towers.

I really need the answers ASAP. I'm going to pass this take home quiz tomorrow at 8:30 AM, Philippine time (GMT +8). 
I suck at math, so I don't know what to do. Thank you so so so much :)
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<U>Answer</U>. The lower tower is 40 m height. The higher tower is 90 m height.


<U>Solution</U>

<pre>
<TABLE>
  <TR>
  <TD>
The scheme is shown in the Figure to the right.
The vertical lines AE and BF represent the towers.
The angle ECF formed by green lines is the right angle.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;


Due to this reason, the triangles ACE and BFC are similar 
  (and their respective vertices are listed in an corresponding order).

From the triangles similarity, we have a proportion {{{abs(AE)/abs(AC)}}} = {{{abs(BC)/abs(BF)}}}, or 
{{{h/60}}} = {{{60/H}}}.   (1)

In addition to it, from the condition we have these equalities:

h = {{{120*tan(beta)}}}  and  H = {{{120*tan(alpha)}}} = {{{120*tan(2*beta)}}}.      (2)
 </TD>
  <TD>
{{{drawing( 350, 250,  -7, 7, -1, 10, 
            line( -6, 0, 6, 0), 
            line( -6, 0, -6, 4),
            line(  6, 0,  6, 9),

            locate(-6.2, -0.2, A),
            locate( 5.8, -0.2, B),
            locate(-0.2, -0.2, C),

            locate(-6.2,  4.9, E),
            locate( 5.8,  9.9, F),

            locate(-6.6,  2.4, h),
            locate( 6.2,  4.9, H),

            locate(-3.8, -0.2, 60_m),
            locate( 2.6, -0.2, 60_m),

        red(line( -6, 0,  6, 9)),
        red(line ( 6, 0, -6, 4)),

            arc ( -6, 0, 2.0, 2.0, 320, 360),
            arc (  6, 0, 2.6, 2.6, 180, 200),

      green(line(  0, 0,  6, 9)),
      green(line ( 0, 0, -6, 4)),

            locate(-4.9, 0.9, alpha=2*beta),
            locate( 3.9, 0.7, beta)

)}}}
 </TD>
 </TR>
</TABLE>

Substituting (2) into (1) we get this equation:  {{{(120*tan(beta))/60}}} = {{{60/(120*tan(2*beta))}}}. 

After canceling by 60, you will get {{{2*tan(beta)}}} = {{{1/(2*tan(2*beta))}}},  or  {{{4*tan(beta)}}} = {{{1/tan(2*beta)}}}.   (3)

Replacing here {{{tan(2*beta)}}} = {{{2*tan(beta)/(1-tan^2(beta))}}},  you will get the equation

{{{4*tan(beta)}}} = {{{(1-tan^2(beta))/(2*tan(beta))}}}.  (4)

To solve it, introduce new variable x = {{{tan(beta)}}}.  You will get 

4x = {{{(1-x^2)/(2x)}}}  --->  {{{8x^2}}} = {{{1 - x^2}}}  --->  {{{9x^2}}} = {{{1}}}  --->  x = {{{1/3}}}.

Thus  {{{tan(beta)}}} = {{{1/3}}},  and you just have   h = {{{120*tan(beta)}}} = {{{120*(1/3)}}} = 40 m.

Next, {{{tan(2*beta)}}} = {{{2*tan(beta)/(1-tan^2(beta))}}} = . . . = {{{3/4}}},  and you finally get 

H = {{{120*tan(2*beta)}}} = {{{120*(3/4)}}} = 90 m.
</pre>