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If in a triangle #ABC# ,#(b+c)/11=(c+a)/12=(a+b)/13# then prove that #cosA/7=cosB/19=cosC/25#

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Answer 1 · 2018-06-05T06:21:29

Given
#InDeltaABC ,(b+c)/11=(c+a)/12=(a+b)/13#

Let

#(b+c)/11=(c+a)/12=(a+b)/13#

#=>(b+c)/11=(c+a)/12=(a+b)/13=(2(a+b+c))/((11+12+13)#

#=>(b+c)/11=(c+a)/12=(a+b)/13=(a+b+c)/18#

#=>a/7=b/6=c/5=k(say)#

So

#cosA=(b^2+c^2-a^2)/(2bc)#

#=>cosA=(36k^2+25k^2-49k^2)/(2*6*5k^2)=1/5#

#cosB=(c^2+a^2-b^2)/(2ca)#

#=>cosB=(25k^2+49k^2-36k^2)/(2*5*7k^2)=19/35#

#cosC=(a^2+b^2-c^2)/(2ab)#

#=>cosC=(49k^2+36k^2-25k^2)/(2*7*6k^2)=5/7#

Now

#cosA:cosB:cosC=1/5:19/35:5/7=7:19:25#

#=>cosA/7=cosB/19=cosC/25#