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

1. ganeshie8

|dw:1443868814415:dw|

2. 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

3. imqwerty

will it not depend upon the side length of square?

4. 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.?

5. ganeshie8

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

6. imqwerty

nope

7. ganeshie8

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

8. 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

9. imqwerty

:)

10. imqwerty

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

11. imqwerty

i mean that regular polygon nd all..

12. 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...

13. ganeshie8

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

14. 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)))

15. amistre64

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

16. amistre64

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

17. amistre64

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

18. 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

19. amistre64

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

20. amistre64

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

21. 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

22. imqwerty

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

23. ganeshie8

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

24. 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.

25. 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.

26. ganeshie8

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

27. ganeshie8

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

28. Empty

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