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Find sin \angle ACB.

 

 Dec 29, 2023
 #1
avatar+129850 
+1

Law of Cosines

 

6^2  = 7^2  + 5^2  - 2(7)(5)cosACB

 

[ 6^2 - 7^2  - 5^2 ] / [ -2 (7)(5) ] =  cos ACB

 

-38 / -70  = cos ACB  =  19/35

 

sin ACB =  sqrt [ 1 - (19/35)^2 ]  =  sqrt  [ 1 - 361/1225]  = sqrt [ 864]/ 35 = 12sqrt(6) / 35  

 

 

cool cool cool

 Dec 29, 2023

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