Question

In $\Delta ABC$ if $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$, then $\cos C$ is equal to

• A. $\frac{5}{7}$
• B. $\frac{7}{5}$
• C. $\frac{16}{17}$
• D. $\frac{17}{36}$

Answer 1

Answer: (A)

$\frac{5}{7}$

Let $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=k$
(i) $b+c=11k$
(ii) $c+a=12k$
(iii) $a+b=13k$
we have, (ii)-(i)
(iv) $a-b=k$
On solving equation, (iii) and (iv) we have,
$a=7k,b=6k,c=5k$
Now, $\cos C=\frac{a^2+b^2-c^2}{2ab}\cos C=\frac{49k^2+36k^2-25k^2}{84k^2}\cos C=\frac{60}{84}\cos C=\frac{5}{7}$