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Mathematics 25 Online
OpenStudy (trojanpoem):

Engineering Drawing.

OpenStudy (trojanpoem):

OpenStudy (trojanpoem):

@BAdhi

OpenStudy (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

OpenStudy (badhi):

OpenStudy (trojanpoem):

But we have 2 unknown sides

OpenStudy (trojanpoem):

Oh, 20 + R and 120 + R

OpenStudy (trojanpoem):

(100)^2 + ( 20 + R)^2 = (120 + R)^2

OpenStudy (trojanpoem):

How about R3

OpenStudy (badhi):

two sides and an angle ;)

OpenStudy (trojanpoem):

(30 + R3) * 0.5 = R3 15 + 0.5R3 = R3 0.5 R3 = 15 R3 = 30 ?

OpenStudy (badhi):

I think so

OpenStudy (badhi):

Find x by looking at the complete 120 line (i hope you can find it easily) ;)

OpenStudy (trojanpoem):

I am dumb.

OpenStudy (trojanpoem):

x= 20 ?

OpenStudy (badhi):

sum of these three will equal to????

OpenStudy (trojanpoem):

30 + 10 + x = 60 x = 20 ? what am I mistaken in ?

OpenStudy (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: \[x = {R_3} = \frac{{80}}{{\sqrt 3 + 1}}\]

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