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If a=b*cos (2pi)/3=c*cos (4pi)/3, then the value of (ab+bc+ca) is ?
Solution :
a=b*cos(2pi)/3 then (3*a)/cos(2pi)=b. Since cos(2pi)=1 ==>(3*a)=b
a=c*cos(4pi)/3 then (3*a)/cos(4pi)=c. Since cos(4pi)=1 ==>(3*a)=c
then (ab+bc+ca)=((3a)(a)+(3a)(3a)+(a)(3a))==>((3a^2)+(9a^2)+(3a^2))=(15*a^2) Ans
Hi Arnab,
cos(2pi/3) = cos[ pi - pi/3 ] = -cos(pi/3) = -1/2
And cos(4pi/3) = cos(pi + pi/3) = -cos(pi/3) = -1/2
So a = -b/2 = -c/2
So b = -2a, c = -2a
ab+bc+ca = a(-2a)+(-2a)(-2a)+(-2a)a = 4a^2 - 4a^2 = 0
Hence the expression is 0.
Hope that helps.
Best Regards,
Ashwin (IIT Madras).
Sorry,
I wrongly interpreted th quetion as a=b*(cos (2pi))/3=c*(cos (4pi))/3, then the value of (ab+bc+ca).
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