A) 5/7
B) 1/5
C) 35/19
D) 19/35
Correct Answer: B
Solution :
Let\[\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=K\] in a triangle ABC. \[\Rightarrow \]\[b+c=11K,c+a=12K,a+b=13K\] On solving these, we get \[a=7K,b=6K,c=5K\]Now we know, \[\cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\] \[=\frac{36{{K}^{2}}+25{{K}^{2}}-49{{K}^{2}}}{2\left( 6K \right)\left( 5K \right)}=\frac{1}{5}\]
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