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Please tell above answer quickly and as soon as possible...........
 
one year ago

Answers : (1)

Arun
24739 Points
							
OA, OB, OC are all same length R so divide by R for unit vectors. 


Construct any ΔA’B’C’ with B’A’||OC, A’C’||OB, C’B’||OA then A’=π−2A, B’=π−2B, C’=π−2C 

( (∠ at centre) = 2(∠ at circumference), so ∠BOC=2A etc. then supplementary ∠s )  

By Sine Rule : |C’B’|/sin(π−2A) = |A’C’|/sin(π−2B) = |B’A’|/sin(π−2C) = k  

Hence |C’B’|=ksin(2A), |A’C’|=ksin(2B), |B’A’|=ksin(2C)  

∴ C’B’ + B’A’ + A’C’ = ksin(2A)OA/R + ksin(2B)OB/R + ksin(2C)OC/R  

But LHS is zero and so sin(2A)OA + sin(2B)OB + sin(2C)OC = 0 
 
one year ago
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