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In any triangle ABC b^2 sin 2C+c^2 sin 2...

In any triangle `ABC b^2 sin 2C+c^2 sin 2B` is equal to

A

`Delta`

B

`2Delta`

C

`3Delta`

D

`4Delta`

Text Solution

Verified by Experts

The correct Answer is:
D

`c^(2)sin2B+b^(2)sin2C`
`=c^(2)(2sinBcosB)+b^(2)(2sinC cosC)`
`=2c^(2)((2Delta)/(ac)"cos"B)+2b^(2)((2Delta)/(ab)"cos"C)`
`=4Delta((c cosB+b cosC)/(a))=4Delta((a)/(a))=4Delta`
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