
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
232.8k+ 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.
Recently Updated Pages
Geometry of Complex Numbers Explained

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

Electricity and Magnetism Explained: Key Concepts & Applications

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions For Class 11 Maths Chapter 12 Limits and Derivatives (2025-26)

NCERT Solutions For Class 11 Maths Chapter 10 Conic Sections (2025-26)

Understanding the Electric Field of a Uniformly Charged Ring

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Derivation of Equation of Trajectory Explained for Students

