Question

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

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

Answer 1

Answer: (B)

$\frac { 5 } { 7 }$

Answer 2

$\frac { b + c } { 11 } = \frac { c + a } { 12 } = \frac { a + b } { 13 }$ = $\lambda$ (Let)

$\therefore$ $b + c = 11 \lambda$ ......(i)

$c + a = 12 \lambda$ .......(ii)

and $a + b = 13 \lambda$ .......(iii)

From (i) + (ii) + (iii), $2 ( a + b + c ) = 36 \lambda$

$\therefore$ $a + b + c = 18 \lambda$ .......(iv)

Now subtract (i), (ii) and (iii) from (iv), $a = 7 \lambda , b = 6 \lambda , c = 5 \lambda$.

Now$\cos C = \frac { a ^ { 2 } + b ^ { 2 } - c ^ { 2 } } { 2 a b }$

= $\frac { ( 7 \lambda ) ^ { 2 } + ( 6 \lambda ) ^ { 2 } - ( 5 \lambda ) ^ { 2 } } { 2.7 \lambda .6 \lambda } = \frac { 49 \lambda ^ { 2 } + 36 \lambda ^ { 2 } - 25 \lambda ^ { 2 } } { 84 \lambda ^ { 2 } }$= $\frac { 60 \lambda ^ { 2 } } { 84 \lambda ^ { 2 } } = \frac { 5 } { 7 }$