Matematik

Help to calculate the height of a cliff in Val d'Isère

03. maj 2014 af freeskier (Slettet) - Niveau: Universitet/Videregående

Hi there all you wise geometricians – I really need your help!! :-)

I want to measure the immense cliff my cousin fell from while skiing back when we were young in Val d’Isère, and now I’ve finally had the chance to measure some of the information to try triangulating my way to the height of the cliff – but I might be still be missing information – or I’m just not wise enough…

This is the information I’ve got – please help me calculate the side “a” (see the illustration and picture)

c = 146 meters
b = 111 meters

As it was snowing and I could see nothing on my phone (angle meter app), I held my ski pole in front of me and measured the distance by eye as well:

a = 96 cm (height of my pole corresponding to the height of the cliff, when held out from my stretched arm (c)
c = 78 cm (the length from my eye to the ski pole (my stretched arm (a))

The two triangles somehow fit together, and I’m hoping to be able to calculate the rest of the triangle in meters (I want to find the side “a” in meters)

Thank you very much in advance for your help (intermediate calculations are appreciated :-)

Tomas :-)

Vedhæftet fil: Illustration and picture.zip

Svar #1
03. maj 2014 af freeskier (Slettet)

Sorry I forgot to mention that the angle "A" represents my point of view in both measurements in meters and cm. It was a little difficult to draw on the picture, but A is supposed to be my eyes! ;-)

One more thing - if this calculation is not possible due to minimum 1 missing information, we could accept the approximation that the angle "C" is a right triangle...
In that case I would like help to the assumption of how much the side "a" changes as a function of the changes in degrees of the angle "C" away from 90 degrees...! :-P

Svar #2
03. maj 2014 af freeskier (Slettet)

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Svar #3
03. maj 2014 af SuneChr

Hey.
The two triangles fit together, - you're right.
We have the relation

$\frac{0,96}{a}=\frac{0,78}{146}$
the same as

$a=\frac{0,96\cdot 146}{0,78}$

# 1 post scriptum
Use the cos-relation
c² = a² + b² - 2ab·cos C

Svar #4
03. maj 2014 af freeskier (Slettet)

Thanks - that looks really obvious! :-D
BUT... Apparently one doesn't seem to be able to draw these kind of conclusions :-(
...because my cousin didn't fall a 180 meters...!!
My best guess is that the cliff (which i vertical - an information I forgot to mention) is between 40-60 meters...

Svar #5
03. maj 2014 af freeskier (Slettet)

So I get that

8995 = a² - 222a · cos(C)

then what...? :-o

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Svar #6
03. maj 2014 af SuneChr

The cos-relation from # 3 is the same as

$a^{2}-\left ( 2b\cdot \cos C \right )\cdot a+\left ( b^{2}-c^{2} \right )=0$ then we have

$a=\frac{2b\cos C\pm \sqrt{\left ( 2b\cos C \right )^{2}-4\cdot \left (b ^{2}-c^{2} \right )}}{2}$

Svar #7
03. maj 2014 af freeskier (Slettet)

So from that I've calculated in the attached Excel (taken out all results a > 100 meters) - is it then just my best guess regarding the approximated angle of C?

Or are there any ways to say more accurately which is more likely - e.g. where the two lines in the graph cross (even though 99 meters seem alot more than expected!!!)??

Vedhæftet fil:Mortens cliff.xlsx

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Svar #8
03. maj 2014 af SuneChr

Can it help?
SP 0305141438.PNG

Vedhæftet fil:SP 0305141438.PNG

Svar #9
03. maj 2014 af freeskier (Slettet)

That's really cool - thanks alot!!! :-D
Nice detail with the picture :-D
I guess I have to inspect that C-angle a little closer! ;-)
Which programme did you use to draw the graph?

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Svar #10
03. maj 2014 af SuneChr

The sky is the limit for graphic design.
Excel 2007 version.

Svar #11
03. maj 2014 af freeskier (Slettet)

Haha - nice :-D

Would you attach your Excel so I can also learn how to set it up in Excel? :-)

And thanks again alot - I've been wondering about this for many years!!
Next time I'm there, I WILL measure that A-angle presisely!! ;-)

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Svar #12
03. maj 2014 af SuneChr

Ohh, I just deleted what I've done in Excel. It is now very easy. Try some different curves with little graphics. You can never do anything wrong or corrupted. I have learned it all by experiment and experiment and .......

Svar #13
03. maj 2014 af freeskier (Slettet)

Ok, so the numbers in my Excel seem correct? I just have to play with the graph type? :-)

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Svar #14
03. maj 2014 af SuneChr

I would like, later, return to the layout of the design. Use vertical columns for C and a.
In the same spreadsheet b and c also can easily changed, - without having to do it all over again.

Svar #15
03. maj 2014 af freeskier (Slettet)

ok - thanks :-)

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Svar #16
03. maj 2014 af SuneChr

http://cossincalc.com/

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