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Question

In triangle ABC,a=3,b=4 and c=5 .Then find the value of sinA+sin2B+sin3C

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Solution

=s(sa)(sb)(sc)=6
cosB=a2+c2b22ac=1692530=3220
sinA=2Ra=bc2=2012=53
sinB=ac2=1512=54
sinC=ab2=1212=1
Now sinA+sin2B+sin3C
=sinA+2sinBcosB+3sinC4sin3C=1425

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