A) (19, 7, 25)
B) (5, 12, 13)
C) (7, 19, 25)
D) (3, 4, 5)
Correct Answer: C
Solution :
Let\[\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=t\] \[\Rightarrow \]\[b+c=11t,c+a=12t,a+b=13t\] \[\Rightarrow \]\[a=7t,b=6t,c=5t\] Now, using cosine rule \[\cos A=\frac{36{{t}^{2}}+25{{t}^{2}}-49{{t}^{2}}}{2.30{{t}^{2}}}=\frac{1}{5}\] Similarly, \[\cos B=\frac{19}{35}\]and\[\cos \,C=\frac{5}{7}\] \[\therefore \]\[\alpha :\beta :\gamma =7:19:25\]
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