Trigonometric Equations Question 111
Question: If in a triangle $ ABC $ , $ \frac{\sin A}{4}=\frac{\sin B}{5}=\frac{\sin C}{6} $ , then the value of $ \cos A+\cos B+\cos C $ is equal to
Options:
A) $ \frac{69}{48} $
B) $ \frac{96}{48} $
C) $ \frac{48}{69} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- We have $ \frac{\sin A}{4}=\frac{\sin B}{5}=\frac{\sin C}{6}\Rightarrow \frac{a}{4}=\frac{b}{5}=\frac{c}{6}=\lambda $ (say)
$ \therefore $ $ a=4\lambda ,,b=5\lambda ,,c=6\lambda $ Now $ \cos A+\cos B+\cos C $ = $ \frac{b^{2}+c^{2}-a^{2}}{2bc}+\frac{c^{2}+a^{2}-b^{2}}{2ac}+\frac{a^{2}+b^{2}-c^{2}}{2ab} $ = $ \frac{1}{240{{\lambda }^{3}}}{4\lambda (45{{\lambda }^{2}}+5\lambda (27{{\lambda }^{2}})+6\lambda (5{{\lambda }^{2}})}=\frac{69}{48} $ .
BETA
