TrojanPoem
  • TrojanPoem
Engineering Drawing.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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TrojanPoem
  • TrojanPoem
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TrojanPoem
  • TrojanPoem
@BAdhi
BAdhi
  • BAdhi
you can find R_1 by looking at the triangle created by ( aligning R_1 and 120 radius) and (R_1 and 20 radius on the lower right) and 160

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BAdhi
  • BAdhi
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TrojanPoem
  • TrojanPoem
But we have 2 unknown sides
TrojanPoem
  • TrojanPoem
Oh, 20 + R and 120 + R
TrojanPoem
  • TrojanPoem
(100)^2 + ( 20 + R)^2 = (120 + R)^2
TrojanPoem
  • TrojanPoem
How about R3
BAdhi
  • BAdhi
two sides and an angle ;)
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TrojanPoem
  • TrojanPoem
(30 + R3) * 0.5 = R3 15 + 0.5R3 = R3 0.5 R3 = 15 R3 = 30 ?
BAdhi
  • BAdhi
I think so
BAdhi
  • BAdhi
Find x by looking at the complete 120 line (i hope you can find it easily) ;)
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TrojanPoem
  • TrojanPoem
I am dumb.
TrojanPoem
  • TrojanPoem
x= 20 ?
BAdhi
  • BAdhi
sum of these three will equal to????
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TrojanPoem
  • TrojanPoem
30 + 10 + x = 60 x = 20 ? what am I mistaken in ?
Michele_Laino
  • Michele_Laino
for \(R_3\) I got this result: |dw:1444412934196:dw| we have: \(L=80 \tan(30) \) so we can write this proportion: \[\frac{{80 - x}}{x} = \frac{{80}}{L}\] from which I get: \[x = {R_3} = \frac{{80}}{{\sqrt 3 + 1}}\]
Michele_Laino
  • Michele_Laino
for \(R_2\) I got this: |dw:1444413551298:dw| so, if we apply the theorem of Pitagora, we have to solve this equation: \[{\left( {30 + x} \right)^2} = {50^2} + {\left( {40 - x} \right)^2}\]
TrojanPoem
  • TrojanPoem
@Michele_Laino , can't we use some drawing operations to find the center point ?
TrojanPoem
  • TrojanPoem
With more than 3 unknown centers , It would be difficult to think. Estimated time is 13 min for each figure.
Michele_Laino
  • Michele_Laino
from my experience, using AutoCAD, all of your arcs will made automatically, we have to use the "grip" at each points, showed by AutoCAD
Michele_Laino
  • Michele_Laino
I recommend you to use a realease at least AutoCAD 2000, not AutoCAD LT 2000
TrojanPoem
  • TrojanPoem
My professor said " You all have to master hand drawing before moving to AutoCAD" Yeah, my problem is the unknown center points it uses concepts like |dw:1444413932573:dw| two tangents to the same circle from the same point has the same length and a line from the mid of both will direct to the center
Michele_Laino
  • Michele_Laino
you have to draw the axes of both segments, and the intersection point of such axes is the center of your circle
TrojanPoem
  • TrojanPoem
That's it , if we know the radius of the circle :/ but it's unknown
Michele_Laino
  • Michele_Laino
|dw:1444414133266:dw|
TrojanPoem
  • TrojanPoem
The radius is perpendicular to the tangent line xD Nice
TrojanPoem
  • TrojanPoem
@Michele_Laino , can't we find R_3 with some drawing operation like what you did ?
Michele_Laino
  • Michele_Laino
I'm thinking...
Michele_Laino
  • Michele_Laino
here is my method: |dw:1444415217388:dw| try to draw a circle which is tangent to both lines #1 and #2 contemporarily
TrojanPoem
  • TrojanPoem
|dw:1444415579441:dw| maybe we can say that the two lines are tangents to the circle so the center of the angle of the intersection point of both lines will cut the the 30 degrees line on the circle mid point
TrojanPoem
  • TrojanPoem
How about R_1 ?
Michele_Laino
  • Michele_Laino
I'm thinking...
TrojanPoem
  • TrojanPoem

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