Engineering Drawing.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Engineering Drawing.

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

1 Attachment
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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question