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**Psalmycole****Member**- Registered: 2011-12-03
- Posts: 9

What is the angle between two adjacent plans of a equilateral triangular pyramid of length 'a' and the perpendicular height of the pyramid is 'h'?

:cool:*Last edited by Psalmycole (2011-12-18 03:49:36)*

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**Psalmycole****Member**- Registered: 2011-12-03
- Posts: 9

I need a simplified formular for this. If not so explain how i can do it.

*Last edited by Psalmycole (2011-12-18 03:33:39)*

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,447

hi Psalmycole

Welcome to the forum.

Yes I can do that.

see diagram

AB = BC = CA = a

Assume pyramid is 'right' => VFA = 90

VF = h

To find the angle between planes VAB and VBC.

(i) Find where the planes intersect.

answer VB

(ii) D is the midpoint of BC

E is the midpoint of AB

F is the point where the lines CE and AD cross.

F is 1/3 of the way along EC

(iii) Find point G on VB so that CGV = 90

Second diagram shows true shape of VBC

(iv) The required angle is AGC

AG = GC

AC = a

Bob

*Last edited by bob bundy (2011-12-18 20:57:53)*

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**Psalmycole****Member**- Registered: 2011-12-03
- Posts: 9

Hi bob? I did follow your work. The last part is like ok, but how u came to get GC, i didn't get it.

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**Psalmycole****Member**- Registered: 2011-12-03
- Posts: 9

Can you prove me right or wrong when i say: cosAGC =(1-6h^2-2a^2)dvide by (a^2)

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,447

hi Psalmycole

When you want the angle between two planes, you have to imagine you are looking along the line of intersection (VB). The lengths that you see are the lengths of lines in each plane that are at 90 degrees to VB.

So I looked at the triangle VBC. We know the length of BC. I had to calculate VB ( = VC ) from the triangle VFC.

As VBC is isosceles I used cosine in the right angled triangle made by splitting VBC in half.

Then I used BGC = 90 and sines to calculate GC ( = GA as the faces are the same).

The formula for VC is not simple so neither is cos(VBC). Then switching to sine makes the formula very complicated.

I followed it through and I do not get

My expression is much more complicated.

Using your formula and with a = 4 and h = 3.265986 (which will make a regular pyramid = tetrahedron ) I got cos AGC = -5.9375 which is not possible!

I used my formulas and got angle = 70.528779 which is the Wiki value.

How did you get your expression?

I'll check your working.

Bob

*Last edited by bob bundy (2011-12-20 06:44:16)*

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,447

My formula is

It took a page of algebra to get to this.

I'll post it if you ask nicely.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**arieneytio****Member**- Registered: 2011-12-24
- Posts: 1

hi,

i'm arieneytio... try to imagine the equilateral triangle base of the pyramid..

then the measure of one of its side is a. using special angle trigonometry, we can solve that the perpendicular distance from the side of the triangular base to the incenter(or the projection of the tip of the pyramid) will be "a(sqr.rt of 3)/4".

forming a right triangle with the height of the pyramid h and base "a(sqr.rt3)/4", we can now use tangent function to get the angle between the two plane.

angle= arc tan [4h/a(sqr.rt of 3)]

Thanks to the ones who didn't help, it's because of them I solve it solely

----arieneytiow----

Imagination is more important than knowledge.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,447

hi arieneytio

Welcome to the forum.

Am I correct in thinking this is for the angle between one sloping face and the base?

EDIT: The in-centre for an equilateral triangle coincides with the centre of gravity = where the medians cross. This point is 1/3 of the way up from the base of the triangle. So

So the angle netween a sloping side and the base is

I was trying to calculate the angle between two sloping sides.

Bob

*Last edited by bob bundy (2011-12-25 20:42:54)*

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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