In a A B C, if b+c 11 =c+a 12 =a+b 13 , then C= - Vidyadip

MCQ

In a $\triangle A B C$, if $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$, then $\cos C=$

A
$\frac{7}{5}$
✓
$\frac{5}{7}$
C
$\frac{17}{16}$
D
$\frac{16}{17}$

✓

Answer

Correct option: B.

$\frac{5}{7}$

(B) Let $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=k$
$\therefore b + c =11 k$ ...(i)
$c + a =12 k$ ...(ii)
and $a + b =13 k$ ...(iii)
From (i) + (ii) + (iii), $2(a+b+c)=36 k$
$\therefore \quad a+b+c=18 k$ ...(iv)
Now, (iv) - (i) gives, $a =7 k$
(iv) - (ii) gives, $b =6 k$
(iv) - (iii) gives, $c =5 k$
Now,
$\cos C =\frac{ a ^2+ b ^2- c ^2}{2 ab }=\frac{(7 k )^2+(6 k )^2-(5 k )^2}{2 \times(7 k ) \times(6 k )}$
$=\frac{49 k ^2+36 k ^2-25 k ^2}{84 k ^2}=\frac{60 k ^2}{84 k ^2}=\frac{5}{7}$

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