wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

In any triangle ABC, prove the following:

a2sin(BC)sinA+b2sin(CA)sinB+c2sin(AB)sinC=0

Open in App
Solution

Given: ABC is a triangle.

According to sine rule

asinA=bsinB=csinC=k ...(i)

We have to prove:

a2sin(BC)sinA+b2sin(CA)sinB+c2sin(AB)sinC=0

LHS=a2sin(BC)sinA+b2sin(CA)sinB+c2sin(AB)sinC

Putting the values from equation (i)

LHS=k2sin2Asin(BC)sinA+k2sin2Bsin(CA)sinB+k2sin2Csin(AB)sinC

=k2sinAsin(BC)+k2sinBsin(CA)+k2sinCsin(AB)

=k2[sinA(sinBcosCsinCcosB)+sinB(sinCcosAsinAcosC)+sinC(sinAcosBsinBcosA)]

=k2(sinAsinBcosCsinAsinCcosB+sinBsinCcosAsinAsinBcosC+sinAsinCcosBsinBsinCcosA)

= 0

= R.H.S

Hence, proved.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon