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 February 20th, 2017, 08:17 AM #1 Newbie   Joined: Dec 2016 From: Latvia Posts: 11 Thanks: 0 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
 February 20th, 2017, 08:43 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,608 Thanks: 1291 the longest side of a right triangle is called a hypotenuse. for the triangle image ... is the second diagram a tetrahedron ... something like this?
February 20th, 2017, 09:25 AM   #3
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Quote:
 Originally Posted by skeeter 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.

February 20th, 2017, 01:30 PM   #4
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 As for first one how did you got results ...

February 20th, 2017, 01:52 PM   #5
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 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, $AB = \sqrt{10^2+(6\sqrt{3})^2} = 4\sqrt{13}$

 February 21st, 2017, 05:41 AM #6 Newbie   Joined: Dec 2016 From: Latvia Posts: 11 Thanks: 0 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.
 February 21st, 2017, 06:38 AM #7 Math Team   Joined: Jul 2011 From: Texas Posts: 2,608 Thanks: 1291 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.
February 21st, 2017, 06:40 AM   #8
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 Originally Posted by skeeter
So how did you get x to be 2√3?

February 21st, 2017, 06:48 AM   #9
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 Originally Posted by skeeter 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.

 February 21st, 2017, 06:49 AM #10 Math Team   Joined: Jul 2011 From: Texas Posts: 2,608 Thanks: 1291 $\sqrt{3} \cdot x = 6$ $x = \dfrac{6}{\sqrt{3}}$ 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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