Trigonometric Equations Question 282
Question: In a $ \Delta ABC, $ if $ \frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13} $ , then $ \cos C= $
[Karnataka CET 2003]
Options:
A) $ \frac{7}{5} $
B) $ \frac{5}{7} $
C) $ \frac{17}{36} $
D) $ \frac{16}{17} $
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Answer:
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} $ .
BETA
