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 #1
avatar+448 
-3
Mar 20, 2019
 #1
avatar+26396 
+4

2013 NS 29

 

1.

cos-theorem:

b2=d2+(ac)22d(ac)cos(L)652=602+58212058cos(L)cos(L)=602+58265212058cos(L)=27396960cos(L)=0.39353448276L=arccos(0.39353448276)L=66.8253951955sin(L)=0.91930985575

 

2.

cos-theorem:

d2=b2+(ac)22b(ac)cos(K)602=652+58213058cos(K)cos(K)=652+58260213058cos(K)=39897540cos(K)=0.52904509284K=arccos(0.52904509284)K=58.0590417221sin(K)=0.84859371300

 

1.sin(K)=rx or r=xsin(K)2.sin(L)=r80x|r=xsin(K)sin(L)=xsin(K)80x(80x)sin(L)=xsin(K)80sin(L)xsin(L)=xsin(K)xsin(L)+xsin(K)=80sin(L)x(sin(L)+sin(K))=80sin(L)x=80sin(L)sin(L)+sin(K)x=800.919309855750.91930985575+0.84859371300x=800.52x=41.6x=41 35

 

The length of segment KA is 41 35

 

laugh

Mar 20, 2019

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