Pyramid

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January 16th 2013, 01:31 AM
Imonars

Pyramid

Hello guys, me and my friends are desperate.. We can't solve this, any ideas? Thanks!
http://i45.tinypic.com/140hx1t.png
January 16th 2013, 02:44 AM
emakarov

Re: Pyramid

Introduce two notations and write two equations for the tangents of the angles shown.
January 16th 2013, 02:49 AM
Imonars

Re: Pyramid

Thanks for the reply
We have tried to find x , h and then solve for h
tan (40.3) = h / x + 200
tan (40.27) = h / x + 100
then we found x and solved for h but we got numbers like 0.525 which seem impossible to represent a high
January 16th 2013, 02:59 AM
emakarov

Re: Pyramid

First, it's 46.27, not 40.27. Second, your equations are wrong because of the order of operations. Last, make sure that the tangent function interprets its arguments as given in degrees, not radians. The correct system gives the height between 440 and 470 feet.
January 16th 2013, 03:07 AM
Imonars

Re: Pyramid

Sorry about the mistake, we have rearranged the equation to be: (x + 100)tan(46.27) = (x + 200)tan(40.3) we are trying to find the height
January 16th 2013, 03:14 AM
emakarov

Re: Pyramid

Yes, this gives x(tan(46.27) - tan(40.3)) = 200 * tan(40.3) - 100 * tan(46.27), i.e.,

$x=\frac{200\tan40.3 - 100\tan46.27}{\tan46.27 - \tan40.3}\approx329.87$

Then h = (x + 100) * tan(46.27).
January 16th 2013, 03:21 AM
Imonars

Re: Pyramid

h = 449.3 :)
Thank you we got it!!
January 16th 2013, 03:30 AM
emakarov

Re: Pyramid

Strictly speaking, if we round the height to one decimal digit, it should be 449.4 because the following digit is 6.
January 16th 2013, 06:01 AM
Soroban

Re: Pyramid

Hello, Imonars!

I would set it up like this:

Code:

o C * * | * * | * * | * * |h * * | * β * α | B o - - - - - - o - - - - - - o D : 100 A x :

A man stands at $A$ and sights the top of the pyramid $C$ .
The angle of elevation is $\alpha = 46.27^o.$

He moves 100 feet away to point $B\!:\;AB = 100.$
The angle of elevation is $\beta = 40.3^o.$

The height of the pyramid is: $h = CD.$
Let $x = AD.$

In $\Delta CDA\!:\;\tan\alpha \,=\, \frac{h}{x} \quad\Rightarrow\quad x \,=\, \frac{h}{\tan\alpha}$ .[1]

In $\Delta CDB\!:\;\tan\beta \,=\, \frac{h}{x+100} \quad\Rightarrow\quad x \,=\,\frac{h}{\tan\beta} - 100$ .[2]

Equate [2] and [1]: . $\frac{h}{\tan\beta} - 100 \;=\;\frac{h}{\tan\alpha}$

Multipy by $\tan\alpha\tan\beta\!:\;h\tan\alpha - 100\tan\alpha\tan\beta \:=\:h\tan\beta$

. . $h\tan\alpha - h\tan\beta \:=\:100\tan\alpha\tan\beta$

. . $h(\tan\alpha - \tan\beta) \:=\:100\tan\alpha\tan\beta$

. . . . . . . . . . . . $h \:=\:\frac{100\tan\alpha\tan\beta}{\tan\alpha - \tan\beta}$

Therefore: . $h \;=\;\frac{100\tan46.27^o\tan40.3^o}{\tan46.27^o - \tan40.3^o} \;=\; 449.364461\hdots$
January 16th 2013, 06:05 AM
Imonars

Re: Pyramid

That's how we got it, we handed in the paper 2 hours ago, the work is the same :) Thanks a alot