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Question

In ABC, asin(BC)+bsin(CA)+csin(AB)=

A
0
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B
a+b+c
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C
a2+b2+c2
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D
2(a2+b2+c2)
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Solution

The correct option is A 0
asin(Bc)+bsin(cA)+csin(AB)
Sin Rule sinAa=sinBsinCk
So write sinA=aK,sinB=bk,sinC=ck
As we know sin(AB)=sinAcosBcosAsinB
Put value of sinA and sinBsin(AB)=akcosBbKcosA=K(2cosBbcosA)
similarity for sin(cA)=k(ccosAacosC)
sin(BC)=k(bcosCccosB)
=ak(bcoscccosB)+bk(ccosAacosc)+ck(acosBbcosβ)=k(bccosAbc(θsA)+k(accosBaccosB)+k(abcoscabcosc
=0+0+0=0
asin(BC)+bsin(CA)+csin(AB)=0
A option is correct.

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