A) \[\frac{5}{7}\]
B) \[\frac{7}{5}\]
C) \[\frac{16}{17}\]
D) \[\frac{17}{36}\]
Correct Answer: A
Solution :
Let us suppose \[\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=k\] \[\therefore \] \[b+c=11k\] ... (i) \[c+a=12k\] ... (ii) \[a+b=13k\] ... (iii) Adding (i), (ii) and (iii) \[2\,(a+b+c)=36k\] \[\therefore \] \[a+b+c=18k\] \[\therefore \] Now \[a=18k-(b+c)\] \[=18k-11k\] [from (i)] \[=7k\] \[b=18k-(a+c)\] \[=18k-6k\] \[\Rightarrow \] \[b=6k\] [from (ii)] \[c=18\,k-(a+b)\] \[=18k-13k\] [from (iii)] \[c=5k\] Now by cosine rule we have \[\cos C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}\] Now putting the values of abc in above formula we get \[\cos C=\frac{{{(7k)}^{2}}+{{(6k)}^{2}}-{{(5k)}^{2}}}{2\,(7k)\,(6k)}\] \[=\frac{49{{k}^{2}}+36{{k}^{2}}-25{{k}^{2}}}{84{{k}^{2}}}\] \[=\frac{85-25}{84}\Rightarrow \frac{60}{84}\] \[=\frac{5}{7}\]
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