# finding height and angle?

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• February 20th 2011, 09:30 AM
andyboy179
Quote:

Originally Posted by scounged
12.5/24=(12.5*2)/(24*2)=25/48

I know it's rinsed by 2 but why do you times it by 2??
• February 20th 2011, 09:35 AM
scounged
Oh, that's just to get rid of the decimal in the numerator for simplicitys sake.
• February 20th 2011, 09:42 AM
andyboy179
so would this mean the angle VAC is also 58.61?
• February 20th 2011, 09:50 AM
scounged
Yes, I think so.
• February 20th 2011, 10:07 AM
andyboy179
wouldn't you double it? because VCM if one triangle piece and VAC is double the size?
• February 20th 2011, 11:06 AM
scounged
Quote:

Originally Posted by andyboy179
wouldn't you double it? because VCM if one triangle piece and VAC is double the size?

I think you're making this task more difficult than it has to be. Let's look at the pyramid from a side point of view:
Attachment 20878

As you can see $\angle(A)=\angle(VAC)=\angle(VAM)$

Notice that the angles VAC and VAM are equal. This doesn't mean that $\triangle(VAC)=\triangle(VAM)$, which in fact is a false statement (oh, btw $\angle(VAC)=\angle(VCA) and \angle(VCM)$).
Angles have nothing to do with the lengths of sides, it's a method of describing the difference in direction between two lines, regardless of the lengths.
• February 20th 2011, 11:12 AM
andyboy179
VAM is 1 triangle part and VAC is 2 so why wouldn't you times 58.61 by 2??

i can't see how VAM and VAC are equal!

VAM=
Attachment 20880

VAC=

Attachment 20881
• February 20th 2011, 12:18 PM
scounged
I'm not sure if I can explain this in a great way, but I'll try.

We need to see the difference between what an angle is and what a triangle is.
Therefore, I will strip my previous images from the sides h and VC (also the point C), leaving us with only two lines with an angle between them:

http://www.mathhelpforum.com/math-he...5&d=1298234507

In this picture we have two sides (VA and AM), and one angle(VAM). Let's see what happens if we extend the side AM a bit:

http://www.mathhelpforum.com/math-he...5&d=1298234508

In this picture the side AM has been extended. As we can see, the angle VAM has not been affected by this change of length in AM.

Let's now examine what happens if we have a triangle. I'll create a triangle out of my first example by adding the side VM:

http://www.mathhelpforum.com/math-he...5&d=1298235238

Surprisingly enough, the angle VAM remains the same.

Let's now take the triangle VAM and extend the side AM to see what happens:

http://www.mathhelpforum.com/math-he...5&d=1298235732

Let's see what has changed here. The length of the side AM has changed. The length of the side VM has changed. Even the area has changed. The angle VAM however, still remains the same. It seems like the angle VAM doesn't depend on the length of the side AM.

Let's merge the triangles and name point M in the second triangle C, and call the distance of VM h:

http://www.mathhelpforum.com/math-he...5&d=1298231489

In this triangle we can see that the angle VAM is equal to the angle VAC (remember that C is the point M in the second triangle). Why is that? Well, it depends on the fact that the side AM has the same direction as the side AC, despite being of a different length, and an angle measures the difference of the direction of two lines, not the difference of lengths or anything else for that matter.

I hope that this has cleared a few things up, and has helped you understand your problem a little better.
• February 20th 2011, 12:34 PM
andyboy179
Oh I think I understand now, so VAC would be 58.61?
• February 20th 2011, 12:38 PM
andyboy179
Also would the height = 20.49
• February 20th 2011, 01:30 PM
scounged
Yes, you're absolutely right.
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