A) \[\frac{7}{5}\]
B) \[\frac{5}{7}\]
C) \[\frac{17}{36}\]
D) \[\frac{16}{17}\]
Correct Answer: B
Solution :
\[\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=\lambda \,\,\,(\text{Let)}\] \[\therefore b+c=11\lambda \] ?.(i) \[c+a=12\lambda \] ?.(ii) and \[a+b=13\lambda \] ?.(iii) From (i) + (ii) + (iii), 2(a + b + c) = 36l, \[\therefore a+b+c=18\lambda \] ?.(iv) Now, (iv) - (i) gives, \[a=7\lambda \] (iv) - (ii) gives, \[b=6\lambda \] (iv) - (iii) gives, \[c=5\lambda \] Now, \[\cos C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}=\frac{{{(7\lambda )}^{2}}+{{(6\lambda )}^{2}}-{{(5\lambda )}^{2}}}{2\times (7\lambda )\times (6\lambda )}\] \[=\frac{49{{\lambda }^{2}}+36{{\lambda }^{2}}-25{{\lambda }^{2}}}{84{{\lambda }^{2}}}\]\[=\frac{60{{\lambda }^{2}}}{84{{\lambda }^{2}}}\] \[=\frac{5}{7}\].
You need to login to perform this action.
You will be redirected in
3 sec
