Finding the angle between slant face and base of a square pyramid, given its net.

- ganeshie8

Finding the angle between slant face and base of a square pyramid, given its net.

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- ganeshie8

|dw:1443868814415:dw|

- ganeshie8

only given information is the radius as shown is \(R\).
Need to find the angle between slant face of square pyramid and the base

- imqwerty

will it not depend upon the side length of square?

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## More answers

- imqwerty

|dw:1443872022196:dw||dw:1443872073345:dw|we can have different cases..
but can we say that there will be a specific case for it to be a square pyramid ..i think no.?

- ganeshie8

you're assuming below angle is 90 degrees, aren't you ?
|dw:1443887889764:dw|

- imqwerty

nope

- ganeshie8

Ahh, I see... It is given that that angle is 90. Sorry about being ambigious ..

- imqwerty

if that is 90* then -|dw:1443891264419:dw| A,B,C,D,E are the vertices of a regular polygon with side length= side length of square
so angle we wanna find=(180-90/4)/2

- imqwerty

:)

- imqwerty

yes lol i wrote that cyclic thingy cause circles came into my mind suddenly and u knw... :)

- imqwerty

i mean that regular polygon nd all..

- ganeshie8

thats not the angle we're after, we want to find the angle that the slant face of a pyramid makes with the base square...

- ganeshie8

I need to find \(\phi\) in below picture :
|dw:1443892934839:dw|

- amistre64

using vectors ....
b = (a/2,h)
a = (0,a)
dot product; a.b
(a/2,h)
( 0 ,a)
-------
ha
magnitudes
|a| = a
|b| = sqrt(a^2/4 + h^2)
phi = acos(ha/sqrt(a^2(a^2/4 + h^2)))

- amistre64

i spose a/a^2 simplifies tho, but its a thought...

- amistre64

phi = acos(h/sqrt(a^2/4 + h^2))
simpler heheh

- amistre64

atan(2h/a) if we want to use the lengths of h and a

- imqwerty

ok
\[\cos \phi=\frac{ a }{ 2b }\]we need to find a and b in terms of R
\[a=\sqrt{R^2+R^2+2\times R \times R \times \cos(\pi/8)}=>a=\sqrt{R^2(1+\cos(\pi/8))}\]\[a=\sqrt{4R^2\sin^2(\pi/16)}\]
b-.|dw:1443894512597:dw|
u knw a in terms of R u can find b too and then jst plug them in

- amistre64

h^2 + (asqrt(2)/2)^2 = R^2

- amistre64

oh, R is face height? or one of the corner hegiths?

- ganeshie8

Ahh that should work perfectly! let me try it once..
I have a semi circular disc from which I am trying to construct pyramid by dividing it into four parts like this :
|dw:1443894809839:dw|
so R is the radius of that circle

- imqwerty

\[b=\sqrt{R^2-\frac{ 4R^2\sin^2(\pi/16) }{ 4}}=>b=\sqrt{R^2(1-\sin^2(\pi/16)}\]
\[b=Rcos(\pi/16)\]
\[\cos \phi=\frac{ 2Rsin(\pi/16) }{ Rcos(\pi/16)}=>\cos \phi=2\tan(\pi/16)\]

- ganeshie8

Awesome!
\[\phi = \cos^{-1}(2\tan(\pi/16))\]
so it doesn't even depend on R, nice! makes sense intuitively too :)

- Empty

imagine 4 vectors \(\vec r_1\) to \(\vec r_4\) all starting at the origin at the top and poking down |dw:1443894424689:dw|
clearly for any one of these, they have length, \(\vec r_i \cdot \vec r_i = R^2\) there are two other possible combinations of dot products, they can either both lie on the same plane, so \(\vec r_i \cdot \vec r_j = R^2 \cos(\frac{\pi}{4})\) or they will lie across from each other \(\vec r_i \cdot \vec r_k = R^2 \cos(?)\)
Then we have these equations
\[\vec b = \frac{\vec r_1 + \vec r_2}{2}\]
\[\vec h = \frac{\vec r_1 + \vec r_2+\vec r_3 + \vec r_4}{2}\]
So from that triangle in there and using the dot product,
\[\cos^{-1} \left(\frac{\vec h \cdot \vec b}{|h||b|}\right) + \frac{\pi}{2} + \phi = \pi\]
I know m answer is incomplete but I just wanted to throw this fun stuff out there, all that's really left to do is plug and chug, and a lot should simplify down by symmetry.

- Empty

imagine 4 vectors \(\vec r_1\) to \(\vec r_4\) all starting at the origin at the top and poking down Created with Raphaëlr2r1r3originReply Using Drawing
clearly for any one of these, they have length, \(\vec r_i \cdot \vec r_i = R^2\) there are two other possible combinations of dot products, they can either both lie on the same plane, so \(\vec r_i \cdot \vec r_j = R^2 \cos(\frac{\pi}{8})\) or they will lie across from each other \(\vec r_i \cdot \vec r_k = R^2 \cos(?)\)
Then we have these equations
\[\vec b = \frac{\vec r_1 + \vec r_2}{2}\]
\[\vec h = \frac{\vec r_1 + \vec r_2+\vec r_3 + \vec r_4}{2}\]
So from that triangle in there and using the dot product,
\[\cos^{-1} \left(\frac{\vec h \cdot \vec b}{|h||b|}\right) + \frac{\pi}{2} + \phi = \pi\]
I know m answer is incomplete but I just wanted to throw this fun stuff out there, all that's really left to do is plug and chug, and a lot should simplify down by symmetry.

- ganeshie8

Nice, that looks complete to me! both formulas seem to be equivalent.. I'm gonna double check..

- ganeshie8

I think \(h\) should be \( \frac{\vec r_1 + \vec r_2+\vec r_3 + \vec r_4}{\color{Red}{4}}\)

- Empty

Oh whoops, forgot to change that when I copied it over haha xD

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