Question

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 1

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.