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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: 2,750 Thanks: 1400 
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  
Newbie Joined: Dec 2016 From: Latvia Posts: 11 Thanks: 0  Quote:
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  
Math Team Joined: Jul 2011 From: Texas Posts: 2,750 Thanks: 1400  Quote:
 
February 20th, 2017, 01:52 PM  #5  
Math Team Joined: Jul 2011 From: Texas Posts: 2,750 Thanks: 1400  Quote:
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 306090 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: 2,750 Thanks: 1400 
Yes, your background in geometry is limited ... which makes me wonder why you are attempting these problems at all. Remediation is available. Two youtube 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 
Newbie Joined: Dec 2016 From: Latvia Posts: 11 Thanks: 0  
February 21st, 2017, 06:48 AM  #9  
Newbie Joined: Dec 2016 From: Latvia Posts: 11 Thanks: 0  Quote:
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: 2,750 Thanks: 1400 
$\sqrt{3} \cdot x = 6$ $x = \dfrac{6}{\sqrt{3}}$ 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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angles, height, knowing, side, triangle 
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