If in a triangle A B C, (b+c)/11=(c+a)/12=(a+b)/13 , then cos A is equal to

Correct answer: (B) 1 / 5. Let (b+c)/11=(c+a)/12=(a+b)/13=K in a triangle A B C . ⇒ b+c=11 K, c+a=12 K, a+b=13 K On solving these, we get a=7 K, b=6 K, c=5 K Now we know, cos A=(b^2+c^2-a^2)/(2 b c) =(36 K^2+25 K^2-49 K^2)/(2(6 K) 5 K)=1/5

Q34 / 90 · JEE Main 2012 (12 May Online) · Mathematics

If in a triangle $A B C, \frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$, then $\cos \mathrm{A}$ is equal to

A
$5 / 7$
Correct answer
B
$1 / 5$
C
$35 / 19$
D
$19 / 35$

Solution

Let $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=K$ in a triangle $A B C$. $\Rightarrow b+c=11 K, c+a=12 K, a+b=13 K$ On solving these, we get $a=7 K, b=6 K, c=5 K$ Now we know, $$ \begin{aligned} & \cos A=\frac{b^2+c^2-a^2}{2 b c} \\ & =\frac{36 K^2+25 K^2-49 K^2}{2(6 K) 5 K}=\frac{1}{5} \end{aligned} $$

Hence, the correct answer is option: 'B'.

If in a triangle $A B C, \frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$, then $\cos \mathrm{A}$ is equal to

Solution

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