Question

If in $∆ABC$, $\frac{(b+c)}{11}=\frac{(c+a)}{12}=\frac{(a+b)}{13}$, then $\cos A$ is equal to:

• A. $\frac{1}{5}$
• B. $\frac{5}{7}$
• C. $\frac{19}{35}$
• D. None of these

Answer 1

The correct option is A

$\frac{1}{5}$

Explanation for the correct option:

Step 1: Find the value of $a+b+c$.

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

$$\begin{gathered} \Rightarrow b+c=11k...\left(1\right) \\ \Rightarrow c+a=12k...\left(2\right) \\ \Rightarrow a+b=13k...\left(3\right) \end{gathered}$$

By adding $\left(1\right),\left(2\right),\mathrm{and}\left(3\right)$, we get

$\begin{array}{rcl}b+c+c+a+a+b& =& 11k+12k+13k\\ \Rightarrow 2\left(a+b+c\right)& =& 36k\\ \Rightarrow a+b+c& =& 18k\end{array}$

Step 2: Find the value of $a,b,\mathrm{and}c$.

$\begin{array}{rcl}a+b+c& =& 18k\\ & \Rightarrow & 13k+c=18k\\ \Rightarrow c& =& 5k\end{array}$

Similarly

$\begin{array}{rcl}a+b+c& =& 18k\\ & \Rightarrow & a+11k=18k\\ \Rightarrow a& =& 7k\end{array}$

and

$\begin{array}{rcl}a+c+b& =& 18k\\ & \Rightarrow & 12k+b=18k\\ \Rightarrow b& =& 6k\end{array}$

Step 3: Apply cosine rule.

As per Cosine Rule,

$\begin{array}{rcl}2bc\cos A& =& b^2+c^2-a^2\\ \Rightarrow \cos A& =& \frac{b^2+c^2-a^2}{2bc}\\ & =& \frac{36k^2+25k^2-49k^2}{2\times 30k^2}\\ & =& \frac{12k^2}{60k^2}\\ & =& \frac{1}{5}\end{array}$

Hence, option A is correct.