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In a triangle ABC, cos 3A+cos 3B+cos3C=1...

In a triangle ABC, `cos 3A+cos 3B+cos3C=1` and `angleA+angleBltangleC`, then find possible measure of `angleC`.

Text Solution

Verified by Experts

The correct Answer is:
`120^(@)`
`cos3A+cos3B+cos 3C=1`
`rARr 2cos(3(A+B))/(2)cos(3(A-B))/(2)=sin^(2)(3C)/(2)`
`rARr -sin(3C)/(2)cos(3(A-B))/(2)=sin^(2)(3C)/(2)`
`rArr sin(3C)/(2)[cos(3(A-B))/(2)-cos(3(A+B))/(2))0`
`rArr 2sin(3A)/(2)sin(3B)/(2)sin(3C)/(2)=0`
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