
In a $\vartriangle ABC$,$\dfrac{\cos a}{a}=\dfrac{\cos b}{b}=\dfrac{\cos c}{c}$ and the side $a=2$, then the area of the triangle is.
A. $1$
B. $2$
C. $\dfrac{\sqrt{3}}{2}$
D. $\sqrt{3}$
Answer
164.1k+ views
Hint: We will use Law of sine $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$ and determine the value of the sides $a,~b$ and \[c\]. We will then substitute the value of the sides determined in the given equation and evaluate. We will get the values of the angles $A,~B$ and $C$, then we will find the area of the triangle by using formula.
Formula Used: Area of the equatorial triangle is $Area=\dfrac{\sqrt{3}}{4}{{a}^{2}}$, where $a$ is the side of the triangle.
Complete step by step solution: We are given a triangle $\vartriangle ABC$ for which $\dfrac{\cos a}{a}=\dfrac{\cos b}{b}=\dfrac{\cos c}{c}$ and the side $a=2$, and we have to find the area of the triangle.
By using Law of sines property of the triangle we can write,
$a=2R\sin A$……. (i)
$b=2R\sin B$……. (ii)
$c=2R\sin C$……(iii)
We will substitute the equations (i), (ii) and (iii) in given equation $\dfrac{\cos a}{a}=\dfrac{\cos b}{b}=\dfrac{\cos c}{c}$.
$\begin{align}
& \dfrac{\cos a}{2R\sin A}=\dfrac{\cos b}{2R\sin B}=\dfrac{\cos c}{2R\sin C} \\
& \dfrac{\cot A}{2R}=\dfrac{\cot B}{2R}=\dfrac{\cot C}{2R} \\
& \cot A=\cot B=\cot C
\end{align}$
All the angles of a triangle will be equal when each of the angles will be ${{60}^{0}}$ which means that it will be an equatorial triangle hence the sides will be also equal, that is $a=b=c=2$.
We will now find the area of the equatorial triangle.
$\begin{align}
& Area=\dfrac{\sqrt{3}}{4}{{(2)}^{2}} \\
& =\dfrac{\sqrt{3}}{4}\times 4 \\
& =\sqrt{3}\,\,uni{{t}^{2}}
\end{align}$
The area of the $\vartriangle ABC$ such that $\dfrac{\cos a}{a}=\dfrac{\cos b}{b}=\dfrac{\cos c}{c}$ and the side $a=2$ is $\sqrt{3}\,\,uni{{t}^{2}}$. Hence the correct option is (D).
Note: The sine law shows the relation between radius of the circumcircle of the triangle and the sines of the angles of the triangle that is $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$ where $A$ is the angle opposite to side $a$, $B$ is the angle opposite to side $b$, $C$ is the angle opposite to side $c$ and $R$ is the radius of the circumcircle of the triangle.
Formula Used: Area of the equatorial triangle is $Area=\dfrac{\sqrt{3}}{4}{{a}^{2}}$, where $a$ is the side of the triangle.
Complete step by step solution: We are given a triangle $\vartriangle ABC$ for which $\dfrac{\cos a}{a}=\dfrac{\cos b}{b}=\dfrac{\cos c}{c}$ and the side $a=2$, and we have to find the area of the triangle.
By using Law of sines property of the triangle we can write,
$a=2R\sin A$……. (i)
$b=2R\sin B$……. (ii)
$c=2R\sin C$……(iii)
We will substitute the equations (i), (ii) and (iii) in given equation $\dfrac{\cos a}{a}=\dfrac{\cos b}{b}=\dfrac{\cos c}{c}$.
$\begin{align}
& \dfrac{\cos a}{2R\sin A}=\dfrac{\cos b}{2R\sin B}=\dfrac{\cos c}{2R\sin C} \\
& \dfrac{\cot A}{2R}=\dfrac{\cot B}{2R}=\dfrac{\cot C}{2R} \\
& \cot A=\cot B=\cot C
\end{align}$
All the angles of a triangle will be equal when each of the angles will be ${{60}^{0}}$ which means that it will be an equatorial triangle hence the sides will be also equal, that is $a=b=c=2$.
We will now find the area of the equatorial triangle.
$\begin{align}
& Area=\dfrac{\sqrt{3}}{4}{{(2)}^{2}} \\
& =\dfrac{\sqrt{3}}{4}\times 4 \\
& =\sqrt{3}\,\,uni{{t}^{2}}
\end{align}$
The area of the $\vartriangle ABC$ such that $\dfrac{\cos a}{a}=\dfrac{\cos b}{b}=\dfrac{\cos c}{c}$ and the side $a=2$ is $\sqrt{3}\,\,uni{{t}^{2}}$. Hence the correct option is (D).
Note: The sine law shows the relation between radius of the circumcircle of the triangle and the sines of the angles of the triangle that is $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$ where $A$ is the angle opposite to side $a$, $B$ is the angle opposite to side $b$, $C$ is the angle opposite to side $c$ and $R$ is the radius of the circumcircle of the triangle.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

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

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Instantaneous Velocity - Formula based Examples for JEE

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

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

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

JEE Advanced 2025 Notes
