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ganeshie8

  • one year ago

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

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  1. ganeshie8
    • one year ago
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    |dw:1443868814415:dw|

  2. ganeshie8
    • one year ago
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    only given information is the radius as shown is \(R\). Need to find the angle between slant face of square pyramid and the base

  3. imqwerty
    • one year ago
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    will it not depend upon the side length of square?

  4. imqwerty
    • one year ago
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    |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.?

  5. ganeshie8
    • one year ago
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    you're assuming below angle is 90 degrees, aren't you ? |dw:1443887889764:dw|

  6. imqwerty
    • one year ago
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    nope

  7. ganeshie8
    • one year ago
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    Ahh, I see... It is given that that angle is 90. Sorry about being ambigious ..

  8. imqwerty
    • one year ago
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    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

  9. imqwerty
    • one year ago
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    :)

  10. imqwerty
    • one year ago
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    yes lol i wrote that cyclic thingy cause circles came into my mind suddenly and u knw... :)

  11. imqwerty
    • one year ago
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    i mean that regular polygon nd all..

  12. ganeshie8
    • one year ago
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    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...

  13. ganeshie8
    • one year ago
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    I need to find \(\phi\) in below picture : |dw:1443892934839:dw|

  14. amistre64
    • one year ago
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    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)))

  15. amistre64
    • one year ago
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    i spose a/a^2 simplifies tho, but its a thought...

  16. amistre64
    • one year ago
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    phi = acos(h/sqrt(a^2/4 + h^2)) simpler heheh

  17. amistre64
    • one year ago
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    atan(2h/a) if we want to use the lengths of h and a

  18. imqwerty
    • one year ago
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    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

  19. amistre64
    • one year ago
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    h^2 + (asqrt(2)/2)^2 = R^2

  20. amistre64
    • one year ago
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    oh, R is face height? or one of the corner hegiths?

  21. ganeshie8
    • one year ago
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    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

  22. imqwerty
    • one year ago
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    \[b=\sqrt{R^2-\frac{ 4R^2\sin^2(\pi/16) }{ 4}}=>b=\sqrt{R^2(1-\sin^2(\pi/16)}\] \[b=Rcos(\pi/16)\] \[\cos \phi=\frac{ 2Rsin(\pi/16) }{ Rcos(\pi/16)}=>\cos \phi=2\tan(\pi/16)\]

  23. ganeshie8
    • one year ago
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    Awesome! \[\phi = \cos^{-1}(2\tan(\pi/16))\] so it doesn't even depend on R, nice! makes sense intuitively too :)

  24. Empty
    • one year ago
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    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.

  25. Empty
    • one year ago
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    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.

  26. ganeshie8
    • one year ago
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    Nice, that looks complete to me! both formulas seem to be equivalent.. I'm gonna double check..

  27. ganeshie8
    • one year ago
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    I think \(h\) should be \( \frac{\vec r_1 + \vec r_2+\vec r_3 + \vec r_4}{\color{Red}{4}}\)

  28. Empty
    • one year ago
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    Oh whoops, forgot to change that when I copied it over haha xD

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