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February 20th, 2017, 09:17 AM   #1
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Triangle side knowing height and angles and...



First one I need to find long sides which name in English I don't know.
For second one I need to find how far k is from a,b,c and how long are edges.
link of img since it doesn't seem to show http://imgur.com/gallery/q3hal
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February 20th, 2017, 09:43 AM   #2
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the longest side of a right triangle is called a hypotenuse.

for the triangle image ...



is the second diagram a tetrahedron ... something like this?

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February 20th, 2017, 10:25 AM   #3
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Quote:
Originally Posted by skeeter View Post
the longest side of a right triangle is called a hypotenuse.

for the triangle image ...



is the second diagram a tetrahedron ... something like this?

As for first one how did you got results I tried cos,sin and Pythagor theorem. but I got different and wrong results( it's entirely possible that I'm doing something wrong though
As for second it's something like that and I need to know AB and BD while I know that AH is 8 and AE is 10.
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February 20th, 2017, 02:30 PM   #4
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Quote:
As for first one how did you got results ...
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February 20th, 2017, 02:52 PM   #5
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Quote:
I need to know AB and BD while I know that AH is 8 and AE is 10.
note that right triangle AHE has sides 6-8-10, so EH = 6

EC, FD, and GB are medians which intersect at the centroid of the base equilateral triangle ... the intersection point divides each median into a 1:2 ratio, so if EH = 6, then HC = 12 and EC = 18.

Traingle BEC is a 30-60-90 triangle with side ratios shown in the previous post

$EC = 18 \implies EB = 6\sqrt{3} \implies BD = 12\sqrt{3}$

Finally, using Pythagoras on triangle AEB, $AB = \sqrt{10^2+(6\sqrt{3})^2} = 4\sqrt{13}$



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February 21st, 2017, 06:41 AM   #6
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Do you maybe have some site with formulas for what I need, since when I try to do it with little I know about geometry and some formulas I found on internet I get different results, and I can get 45 45 90 triangle to be right if I put x in like this √2*36 but it doesn't work for other one and if I try to get 6 from √3x other numbers are wrong and huge, and as for second just to make sure AD=AC=AB and how would you prove that EH is one third of EC?
Sorry if these are dumb questions, it's just that there is allot of simple things that I haven't learned for one reason or other.
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February 21st, 2017, 07:38 AM   #7
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Yes, your background in geometry is limited ... which makes me wonder why you are attempting these problems at all.

Remediation is available. Two you-tube videos to start. First one is about special right triangles; second is about medians of a triangle.





Good luck.
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February 21st, 2017, 07:40 AM   #8
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So how did you get x to be 2√3?
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February 21st, 2017, 07:48 AM   #9
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Quote:
Originally Posted by skeeter View Post
Yes, your background in geometry is limited ... which makes me wonder why you are attempting these problems at all.

Remediation is available. Two you-tube videos to start. First one is about special right triangles; second is about medians of a triangle.





Good luck.
Thanks.
And as for stupid reason, I'm lazy and could get by without studying, I'm trying to catch up, unfortunately haven't got this far.
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February 21st, 2017, 07:49 AM   #10
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$\sqrt{3} \cdot x = 6$

$x = \dfrac{6}{\sqrt{3}}$

rationalize the denominator ...

$\dfrac{6}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3} = 2\sqrt{3}$

Maybe you should also look at this video ...

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