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If in A B C ,A=pi/7,B=(2pi)/7,C=(4pi)/7...

If in ` A B C ,A=pi/7,B=(2pi)/7,C=(4pi)/7` then `a^2+b^2+c^2` must be (a)`R^2` (b) `3R^2` (c) `4R^2` (d) `7R^2`

A

`R^(2)`

B

`3R^(2)`

C

`4R^(2)`

D

`7R^(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
`a^(2) + b^(2) + c^(2) = 4R^(2) (sin^(2) A + sin^(2) B + sin^(2)C)`
`= 2R^(2) (3 - (cos 2A + cos 2B + cos 2C))`
Now, `cos 2A + cos 2B + cos 2C`
`= cos.(2pi)/(7) + cos.(4pi)/(7) + cos.(8pi)/(7)`
`= cos.(2pi)/(7) + cos.(4pi)/(7) + cos.(6pi)/(7)`
`= (sin.(3pi)/(7))/(sin.(pi)/(7)) cos.(4pi)/(7)`
`= (-2 sin.(3pi)/(7) cos.(3pi)/(7))/(2 sin.(pi)/(7))`
`= (-sin.(6pi)/(7))/(2sin.(pi)/(7)) = (-1)/(2)`
`:. a^(2) +b^(2) + c^(2) = 2R^(2) (3-((-1)/(2))) = 7R^(2)`
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