In a triangle `ABC, cos 3A + cos 3B + cos 3C = 1`, then find any one angle.

Options

A. `30^(@)`
B. `60^(@)`
C. `90^(@)`
D. `120^(@)`

Answer

D

Explanation

We have,
`cos3A+cos3B+cos3C=1`
`implies2cos3((A+B)/(2))cos3((A-3)/(2))-cos3(A+B)=1" "[becauseC=pi-(A+B)]`
`implies2cos2((A+3)/(2))cos3((A-B)/(2))-cos^(2)3((A+B)/(2))=0`
`implies2cos3((A+B)/(2)){cos3((A-B)/(2))-cos3((A+B)/(2))}=0`
`impliescos3((A+B)/(2))=0or,orcos3((A-B)/(2))=cos3((A+B)/(2))`
Now, `cos3((A+B)/(2))=0impliesA+B=(pi)/(3)impliesC=(2pi)/(3)`
and, `cos3((A-B)/(2))=cos3((A+B)/(2))impliesB=0,` which is not possible.