Question 1187818
.
An observer 9 meters horizontally away from the tower observes its angle of elevation to be only one half 
as much as the angle of elevation of the same tower when he moves 5 meters nearer towards the tower. 
How high is the tower?
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            It is a good  Trigonometry problem,  and it deserves a detailed explanation.


            See my solution below.  Read it attentively.



<pre>
Let h be the height of the tower.


First position is 9 meters from the tower.  In this position,

    tan(a) = {{{h/9}}},  where "a" is the angle of elevation in this position.    (1)


Next position is  (9-5) = 4 meters from the tower.  In this position,

    tan(b) = {{{h/4}}},  where "b" is the angle of elevation in this position.    (2)


We are given  b = 2a.  Hence,

    tan(b) = tan(2a) = (use the basic Trigonometry formula) = {{{(2*tan(a))/(1-tan^2(a))}}} = 


           = {{{(2*(h/9))/(1 - (h/9)^2)}}} = (simplify) = {{{(2*h*81)/(9*(81-h^2))}}} = {{{(2*h*9)/(81-h^2)}}} = {{{(18h)/(81-h^2)}}}.


Thus from (2) we have THIS EQUATION

    {{{(18h)/(81-h^2)}}} = {{{h/4}}}.


Cancel common factor "h" in both sides; then cross-multiply to get

    18*4 = 81 - h^2

     72  = 81 - h^2

    h^2 = 81 - 72 = 9.


Hence,  h = {{{sqrt(9)}}} = 3.


<U>ANSWER</U>.  The tower height is 3 meters.
</pre>

Solved and thoroughly explained.