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Calc A-C in Equilateral Triangle where A-B = 100mm

Posted on 2013-05-11
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Last Modified: 2013-05-13
given an equilateral triangle where all three sides are 100mm, how does one calculate from any point to the center of the triangle?
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Question by:volking
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by:ozo
ID: 39158859
You can use the Pythagorean theorem
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by:aadih
ID: 39158870
P-Theorem: right-angled triangle only.
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by:ozo
ID: 39158873
There are several right-angled triangles in your diagram
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by:aadih
ID: 39158874
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aadih earned 500 total points
ID: 39158909
So, did you calculate yet?

AC = L/sqrt(3) mm  = 100/sqrt(3) = 57.7 mm.
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by:Paul Sauvé
ID: 39158912
in geometry:
equilateral triangle each angle equals 180/3 deg = 60 deg
in trigonometry:
angle ABC = BAC = 30 deg
cos 30 deg = 50 mm/AC =  0,8660
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by:ozo
ID: 39158914
and cos 30 deg is calculated from
the Pythagorean theorem
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by:Paul Sauvé
ID: 39158920
@ aadih, I didn't know that ;-d
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by:TommySzalapski
ID: 39158946
diagramYou can do it with only the Pythagorean theorem and some Algebra.
Label the other vertex D, and the middle of AB, E (So DE is the height of the triangle).
Note that DEA and DEB are right triangles
Well, clearly DE*DE + BE*BE =  BD*BD
BD = 100
BE = 50
So you can easily get DE
Now DE = CE + CD
From Pythagorean: CB*CB = CE*CE+BE*BE
Since DC = CB you can replace it giving two equations with two unknowns.
Solve that and you'll get the same 100/sqrt(3)
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by:TommySzalapski
ID: 39158953
There's an even better way to do it though.
Note that the angle at D is clearly bisected by ED. Since the angles where the lines bisect the sides of the triangle are right angles, then triangle DEB and BEC are similar triangles.
That means that DB/ED = BC/BE. Then also, BC + CE = DE (since BC=DC)
So again you have two equations and two unknowns. Solving should be simple(ish) from there.
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by:käµfm³d 👽
ID: 39158989
@ozo
There are several right-angled triangles in your diagram
Please correct me if I am mistaken, but there is nothing in the diagram to indicate that those are right angles. I was always told in my geometry classes that you cannot assume an angle is a right angle--it must be clearly denoted. In other words, this triangle shows 3 right triangles:

Right Angles
This triangle may or may not have right triangles:

Unknown Right Angles
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by:TommySzalapski
ID: 39158999
Well, you can't get the answer unless you know how those lines were drawn. If you are given (as it must have been) that those lines bisect either the angles or the sides, then it is trivial to prove that they are right angles. (equilateral is 60 degrees at each angle so you have 30, 60, 90 triangles).
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by:käµfm³d 👽
ID: 39159027
Sure, if you know that the lines bisect either the side or the angle, the you can assume a right angle. As written, there is not enough information to calculate anything. IIRC, you need at least two pieces of information to calculate missing angles and sides when dealing with triangles.

*Edit

Actually, since the OP mentions that C is the center, I think there is some law of equilateral triangles that says that each line drawn from an opposite angle through the center bisects the opposing side. So I withdraw my previous comment (though I cannot quote the law to which I am referring).
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by:aburr
ID: 39159059
by symmetry you can say that the line through the center from a vertex bisects AC and also bisects the 60 degree angel at the vertex. From there you can construct any number of 90 degree triangles.
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by:ozo
ID: 39159111
DEB and BEC are similar triangles.
and you can prove the Pythagorean theorem with similar triangles.
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