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Find the circumradius of triangle $ABC.$

[asy] unitsize(1 cm);pair A, B, C;A = (1,3);B = (0,0);C = (4,0);draw(A--B--C--cycle);label(

 Apr 12, 2022
 #1
avatar+124592 
+1

C = 45°

 

By the Law of Sines

 

8 /sin 45  =  BC / sin 60

 

8 sin 60 / sin 45  = BC =  4sqrt 6

 

And

 

8 /sin 45  = AC /sin75

 

8sin75 / sin 45 = AC  = 4 ( 1 + sqrt 3)

 

Area of triangle = A  =   (1/2)AB *BC *sin75  =  (1/2) (  8) ( 4sqrt 6) sin 75   =  24 + 8sqrt (3)

 

AB * BC * AC  =   8 * [ 4sqrt (6) ]  * [ 4 (1 + sqrt (3) ) ]  =   128 sqrt (6)  ( 1 + sqrt (3) )

 

 

Circumradius  =      [ AB * BC * AC ]  / [  4 A ]  =

 

[   128 sqrt (6)  ( 1 + sqrt (3) ) ]  / [ 4 * ( 24 + 8sqrt (3)) ]    =   4sqrt (2)

 

 

cool cool cool

 Apr 12, 2022
 #2
avatar+114 
-2

Mhmmmmmm

 Apr 12, 2022

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