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Please help with this question.

 Apr 15, 2022
 #1
avatar+13900 
+2

Find the sum of alldistinct of BC.

 

Hello Guest!

 

\(c^2=a^2+b^2-2bc\ cos\ \gamma\\ 4^2=x^2+10^2-2\cdot 10\cdot x\cdot cos\frac{\pi}{6}\\ x^2-20\cdot cos\frac{\pi}{6}\cdot x+100-16\\ x=8.66\pm\sqrt{8.66^2-84}\\ x\in \{\} \)

\(BC\in\{\}\)

laugh  !

 Apr 15, 2022
edited by asinus  Apr 15, 2022
 #2
avatar+124598 
+1

                  B

                          4

C 30°          10                  A

 

 

4^2  = 10^2 + BC^2  -  2 (10)(BC) ( sqrt 3)/2)

 

16 = 100 + BC^2  - (10sqrt 3)BC

 

BC^2 - (10sqrt 3)BC + 84   = 0

 

[ No real solutions for BC     since the discriminant is  negative ] 

 

 

cool cool cool

 Apr 15, 2022
 #3
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0

Thanks for answering Asinus and CPhill!!!

 Apr 15, 2022

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