ganeshie8
  • ganeshie8
Finding the angle between slant face and base of a square pyramid, given its net.
AP Math
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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ganeshie8
  • ganeshie8
|dw:1443868814415:dw|
ganeshie8
  • 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
  • imqwerty
will it not depend upon the side length of square?

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imqwerty
  • 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
  • ganeshie8
you're assuming below angle is 90 degrees, aren't you ? |dw:1443887889764:dw|
imqwerty
  • imqwerty
nope
ganeshie8
  • ganeshie8
Ahh, I see... It is given that that angle is 90. Sorry about being ambigious ..
imqwerty
  • 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
:)
imqwerty
  • imqwerty
yes lol i wrote that cyclic thingy cause circles came into my mind suddenly and u knw... :)
imqwerty
  • imqwerty
i mean that regular polygon nd all..
ganeshie8
  • 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
  • ganeshie8
I need to find \(\phi\) in below picture : |dw:1443892934839:dw|
amistre64
  • 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
  • amistre64
i spose a/a^2 simplifies tho, but its a thought...
amistre64
  • amistre64
phi = acos(h/sqrt(a^2/4 + h^2)) simpler heheh
amistre64
  • amistre64
atan(2h/a) if we want to use the lengths of h and a
imqwerty
  • 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
  • amistre64
h^2 + (asqrt(2)/2)^2 = R^2
amistre64
  • amistre64
oh, R is face height? or one of the corner hegiths?
ganeshie8
  • 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
  • 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
  • ganeshie8
Awesome! \[\phi = \cos^{-1}(2\tan(\pi/16))\] so it doesn't even depend on R, nice! makes sense intuitively too :)
Empty
  • 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
  • Empty
imagine 4 vectors \(\vec r_1\) to \(\vec r_4\) all starting at the origin at the top and poking down Created with Raphaƫlr2r1r3originReply 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
  • ganeshie8
Nice, that looks complete to me! both formulas seem to be equivalent.. I'm gonna double check..
ganeshie8
  • ganeshie8
I think \(h\) should be \( \frac{\vec r_1 + \vec r_2+\vec r_3 + \vec r_4}{\color{Red}{4}}\)
Empty
  • Empty
Oh whoops, forgot to change that when I copied it over haha xD

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