My Math Forum Height of the tower

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 March 28th, 2013, 04:01 AM #1 Member   Joined: Nov 2012 Posts: 61 Thanks: 0 Height of the tower The angles of elevation of the top of a tower from two points distant "s" and "t" from its foot are complementary. Prove that the height of the tower is$\sqrt{st}$.
 March 28th, 2013, 04:54 AM #2 Senior Member   Joined: Feb 2013 Posts: 281 Thanks: 0 Re: Height of the tower You have two right-angled triangles. One side is mutual, the height (h) of the tower. The other side is s and t. The connection between the two perpendicular sides is h/s = tan(alpha) in one triangle and h/t = tan(90-alpha). Try to finish the proof.
 March 28th, 2013, 04:59 AM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Re: Height of the tower Let h be the height of the tower. $s\,=\,\frac{h}{\tan(A)},\,t\,=\,\frac{h}{\tan(90\,-\,A)} \\ st\,=\,\frac{h^2}{\tan(A)\,\cdot\,\tan(90\,-\,A)}$
March 28th, 2013, 05:20 AM   #4
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Re: Height of the tower

Hello, rakmo!

Quote:
 The angles of elevation of a top of a tower from two points with distances $s$ and $t$ from its foot are complementary.[color=beige] .[/color]Prove that height of the tower is$\sqrt{st}$

Code:
                        o P
* *|
*   *a|
*     *  | h
*       *   |
* a       * b  |
o-----------o-----o
Q           R  t  S
: - - -  s  - - - :
$\text{The height of the tower is: }\:h \,=\,PS.$

$\text{From point }Q: \; QS = s,\:\angle PQS = a
\text{From point }R: \; RS = t,\:\angle PRS = b$

$\text{Since }a\text{ and }b\text{ are complementary, }\,\angle RPS= a.$

$\text{In }\Delta PSR:\;\tan\,\! a \,=\, \frac{t}{h}\;\;[1]$

$\text{In }\Delta PSQ:\;\tan\,\! a \,=\, \frac{h}{s}\;\;[2]$

$\text{Equate [1] and [2]: }\;\frac{t}{h} \,=\,\frac{h}{s} \;\;\;\Rightarrow\;\;\; h^2 \,=\,st \;\;\;\Rightarrow\;\;\; h \,=\,\sqrt{st}$

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