In a triangle ABC, #tanA:tanB:tanC = 1:2:3#. Hence #sinA:sinB:sinC# is equal to?

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Answer 1 · 2017-12-12T09:33:46

Given

#tanA:tanB:tanC = 1:2:3#.
Let #tanA=k;tanB=2kand tanC=3k#,where #k!=0#

Also given that A,B and C are three angles of a triangle ABC.

So #A+B+C=pi#

#=>tan(A+B)=tan(pi-C)#

#=>(tanA+tanB)/(1-tanAtanB)=-tanC#
#=>tanA+tanB=-tanC+tanAtanB tanC#

#=>tanA+tanB+tanC=tanAtanBtanC#

#=>k+2k+3k=kxx2kxx3k#

#=>6k=k^3#

As #k!=0#

We get #k^2=1 andk=1#

(#k=-1->#) neglected as it gives more than one obtuse angle in a triangle, which is impossible

Now #sinA=tanA/secA=tanA/sqrt(1+tan^2A)#

#=k/sqrt(1+k^2)=1/sqrt2#

#sinB=tanB/secB=tanB/sqrt(1+tan^2B)#

#=(2k)/sqrt(1+4k^2)=2/sqrt5#

#sinC=tanC/secC=tanC/sqrt(1+tan^2C)#

#=(3k)/sqrt(1+9k^2)=3/sqrt10#

Hence #sinA:sinB:sinC#

#=1/sqrt2:2/sqrt5:3/sqrt10#

#=sqrt5:2sqrt2:3#