
In a triangle$\vartriangle ABC$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
161.7k+ views
Hint: To solve this question, we will assume the given equation to some variable constant and form three equations. Then with the help of all three equations by performing some arithmetical operation we will derive the value of $a,b$and $c$. Then we will substitute values of $a,b$ and $c$in the cosine rule \[{{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos C\]and find the value of angle $\cos C$.
Complete step-by-step solution:
We are given a triangle$\vartriangle ABC $for which $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$ and we have to find the value of angle $\cos C$.
Let us assume that $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=k$,
So we can write,
b+c=11k..........(i)
c+a=12k..........(ii)
a+b=13k..........(iii)
We will now add all the three equations (i),(ii) and (iii),
-$\begin{align}
& b+c+c+a+a+b=11k+12k+13k \\
& 2(a+b+c)=36k \\
& a+b+c=18k..........(iv)
\end{align}$
We will now subtract equation (i) from equation (iv) and simplify the equation to find the value of $a$.
$\begin{align}
& a+b+c-(b+c)=18k-11k \\
& a=7k
\end{align}$
We will now subtract equation (ii) from equation (iv) and simplify the equation to find the value of $b$.
-$\begin{align}
& a+b+c-(c+a)=18k-12k \\
& b=6k
\end{align}$
We will now subtract equation (iii) from equation (iv) and simplify the equation to find the value of $c$.
$\begin{align}
& a+b+c-(a+b)=18k-13k \\
& c=5k
\end{align}$
We will now substitute the value of $a,b$ and $c$ in the cosine rule and then simplify the equation.
\[\begin{align}
& {{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos C \\
& {{(5k)}^{2}}={{(7k)}^{2}}+{{(6k)}^{2}}-2(7k)(6k)\cos C \\
& 25{{k}^{2}}=49{{k}^{2}}+36{{k}^{2}}-84{{k}^{2}}\cos C \\
& 84{{k}^{2}}\cos C=60{{k}^{2}} \\
& \cos C=\frac{5}{7}
\end{align}\]
The triangle $\vartriangle ABC$ for which $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$ , the value of the angle $\cos C$is \[\cos C=\frac{5}{7}\].Hence the correct option is (B).
Note:
There are three cosine rules or Law of cosine with different angles in each of them so we must select the cosine rule according to the question.
We can use any other way to find the value of $a,b$ and $c$by using a substitution method or any other different method.
Complete step-by-step solution:
We are given a triangle$\vartriangle ABC $for which $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$ and we have to find the value of angle $\cos C$.
Let us assume that $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=k$,
So we can write,
b+c=11k..........(i)
c+a=12k..........(ii)
a+b=13k..........(iii)
We will now add all the three equations (i),(ii) and (iii),
-$\begin{align}
& b+c+c+a+a+b=11k+12k+13k \\
& 2(a+b+c)=36k \\
& a+b+c=18k..........(iv)
\end{align}$
We will now subtract equation (i) from equation (iv) and simplify the equation to find the value of $a$.
$\begin{align}
& a+b+c-(b+c)=18k-11k \\
& a=7k
\end{align}$
We will now subtract equation (ii) from equation (iv) and simplify the equation to find the value of $b$.
-$\begin{align}
& a+b+c-(c+a)=18k-12k \\
& b=6k
\end{align}$
We will now subtract equation (iii) from equation (iv) and simplify the equation to find the value of $c$.
$\begin{align}
& a+b+c-(a+b)=18k-13k \\
& c=5k
\end{align}$
We will now substitute the value of $a,b$ and $c$ in the cosine rule and then simplify the equation.
\[\begin{align}
& {{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos C \\
& {{(5k)}^{2}}={{(7k)}^{2}}+{{(6k)}^{2}}-2(7k)(6k)\cos C \\
& 25{{k}^{2}}=49{{k}^{2}}+36{{k}^{2}}-84{{k}^{2}}\cos C \\
& 84{{k}^{2}}\cos C=60{{k}^{2}} \\
& \cos C=\frac{5}{7}
\end{align}\]
The triangle $\vartriangle ABC$ for which $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$ , the value of the angle $\cos C$is \[\cos C=\frac{5}{7}\].Hence the correct option is (B).
Note:
There are three cosine rules or Law of cosine with different angles in each of them so we must select the cosine rule according to the question.
We can use any other way to find the value of $a,b$ and $c$by using a substitution method or any other different method.
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