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ganeshie8
 one year ago
Finding the angle between slant face and base of a square pyramid, given its net.
ganeshie8
 one year ago
Finding the angle between slant face and base of a square pyramid, given its net.

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5dw:1443868814415:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5only given information is the radius as shown is \(R\). Need to find the angle between slant face of square pyramid and the base

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.5will it not depend upon the side length of square?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.5dw:1443872022196:dwdw:1443872073345:dwwe 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
 one year ago
Best ResponseYou've already chosen the best response.5you're assuming below angle is 90 degrees, aren't you ? dw:1443887889764:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5Ahh, I see... It is given that that angle is 90. Sorry about being ambigious ..

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.5if 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=(18090/4)/2

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.5yes lol i wrote that cyclic thingy cause circles came into my mind suddenly and u knw... :)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.5i mean that regular polygon nd all..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5thats 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
 one year ago
Best ResponseYou've already chosen the best response.5I need to find \(\phi\) in below picture : dw:1443892934839:dw

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2using 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
 one year ago
Best ResponseYou've already chosen the best response.2i spose a/a^2 simplifies tho, but its a thought...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2phi = acos(h/sqrt(a^2/4 + h^2)) simpler heheh

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2atan(2h/a) if we want to use the lengths of h and a

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.5ok \[\cos \phi=\frac{ a }{ 2b }\]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
 one year ago
Best ResponseYou've already chosen the best response.2h^2 + (asqrt(2)/2)^2 = R^2

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2oh, R is face height? or one of the corner hegiths?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5Ahh 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
 one year ago
Best ResponseYou've already chosen the best response.5\[b=\sqrt{R^2\frac{ 4R^2\sin^2(\pi/16) }{ 4}}=>b=\sqrt{R^2(1\sin^2(\pi/16)}\] \[b=Rcos(\pi/16)\] \[\cos \phi=\frac{ 2Rsin(\pi/16) }{ Rcos(\pi/16)}=>\cos \phi=2\tan(\pi/16)\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5Awesome! \[\phi = \cos^{1}(2\tan(\pi/16))\] so it doesn't even depend on R, nice! makes sense intuitively too :)

Empty
 one year ago
Best ResponseYou've already chosen the best response.2imagine 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, \[\cos^{1} \left(\frac{\vec h \cdot \vec b}{hb}\right) + \frac{\pi}{2} + \phi = \pi\] 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
 one year ago
Best ResponseYou've already chosen the best response.2imagine 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, \[\cos^{1} \left(\frac{\vec h \cdot \vec b}{hb}\right) + \frac{\pi}{2} + \phi = \pi\] 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
 one year ago
Best ResponseYou've already chosen the best response.5Nice, that looks complete to me! both formulas seem to be equivalent.. I'm gonna double check..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5I think \(h\) should be \( \frac{\vec r_1 + \vec r_2+\vec r_3 + \vec r_4}{\color{Red}{4}}\)

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Oh whoops, forgot to change that when I copied it over haha xD
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