Question 168863
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To begin with, 1st we need to find the ANGLES of the triangle.
Given Ratio ----> 3:4:5 ------>{{{system(A=45^o,B=60^o,+C=75^o,180^0)}}}
Agree? Divide All Angles by 15: {{{45^o/15=highlight(3)}}};{{{60^0/15=highlight(4)}}};{{{75^o/15=highlight(5)}}}
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Let's put those angles in figure so it will be easy to analyze:
{{{drawing(400,400,-.20,5.5,-.50,5,triangle(0,0,5,0,3,3),locate(-.10,0,A),locate(5.10,0,C),locate(3.10,3.40,B),locate(.50,.40,45^o),locate(2.80,2.75,60^o),locate(4.10,.40,75^o),locate(4.25,2,1m))}}} ---> As you see in our drawing, 
{{{BC}}} is the shortest side = 1meter. 
For argument sake, take a protractor and 
draw on your own to have the right angles and measure the shortest.
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Here's the trick, determining the longset side because it seems the other 2 sides are so close. Draw a Line from {{{B}}} to Line {{{AC}}} and mark point {{{D}}}:
{{{drawing(400,400,-.20,5.5,-.50,5,triangle(0,0,5,0,3,3),locate(-.10,0,A),locate(5.10,0,C),locate(3.10,3.40,B),locate(.50,.40,45^o),locate(2.80,2.75,60^o),locate(4.10,.40,75^o),locate(4.25,2,1m),locate(3,-.10,D),red(line(3,3,3,0)),locate(2.50,.40,90^o),locate(3.10,.40,90^o))}}} 
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We can solve for Line {{{BD}}} by Pyth Theorem -->{{{(BC)^2=(BD^2)+(DC)^2}}} -->
{{{(BD)^2=(BC)^2-(DC)^2}}} ---------> EQN 1
*note: {{{BC=1m}}}, and we need to find {{{DC}}}?
By trigo function: {{{Cos75^o=adj/hyp=DC/BC}}}--->{{{DC=(cos75^o)(BC)=(cos75^o)(1)}}} --> {{{DC=0.2588m}}}, Plug in EQN 1:
{{{(BD)^2=1^2-0.2588^2=1-0.066977=(0.93302m)}}}
{{{BD=sqrt(0.93302)=0.966023m}}}
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Now we need to find {{{AD}}}. By Trigo function, {{{Tan45^o=BD/AD}}}
{{{AD=BD/Tan45^o=0.966023/Tan45^o=0.966023m}}}
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So, line {{{AC=AD+DC=0.966023+0.2588=1.2248m}}}
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Now,we find Line {{{AB}}}. We can do that either by Trigo Function or Cosine Law. Let's try both and we'll see if the same.
By Trigo function---> {{{Sin45^o=opp/hyp=BD/AB}}}
{{{AB=BD/Sin45^o=0.966023/Sin45^o=1.36616m}}}
Or By Cosine Law ---> {{{(AB)^2=(AC)^2+(BC)^2-2(AC)(BC)Cos75^o}}}
{{{(AB)^2=1.2248^2+1^2-2(1.2248)(1)Cos75^o}}}
{{{(AB)^2=1.5+1-0.6340=1.866}}}
{{{AB=sqrt(1.866)=1.366m}}}, The same above
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So here are your measurements:
{{{highlight(AB=1.366m)}}} ----------> Longest
{{{BC=1m}}} -------------------------> Shortest
{{{AC=1.2248m}}} --------------------> Middle
{{{drawing(400,400,-.20,5.5,-.50,5,triangle(0,0,5,0,3,3),locate(-.10,0,A),locate(5.10,0,C),locate(3.10,3.40,B),locate(.50,.40,45^o),locate(2.80,2.75,60^o),locate(4.10,.40,75^o),locate(4.25,2,1m),locate(3,-.10,D),red(line(3,3,3,0)),locate(2.50,.40,90^o),locate(3.10,.40,90^o),locate(.80,2,highlight(1.366m)),locate(2.50,-.30,1.2248m))}}} 
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Thank you,
Jojo