Question

The perimeter of $\vartriangle ABC$ is $6$ times the arithmetic mean of the sines of its angles. If the side $a$ is $1$, then the angle $A$ is .
A. $\dfrac{\pi }{6}$
B. $\dfrac{\pi }{3}$
C. $\dfrac{\pi }{2}$
D. $\pi $

Answer 1

Hint: We will first find the perimeter of the triangle and arithmetic mean of the sines of the angles of the triangle by using their respective formulas. After this we will write both the values as per the condition given in the question and form an equation. By using the Law of sines we will derive the value of the length of the sides $a,b,c$ and substitute it in the equation and find the value of angle $A$.

Formula Used: $Arithmetic\,\,mean=\dfrac{Sum\,\,of\,observations}{Number\,\,of\,observations.}$

Complete step by step solution: We are given a triangle $\vartriangle ABC$ of which the perimeter is $6$ times the arithmetic mean of the sines of its angles. One side of that triangle $a$ is $1$ then we have to find the angle $A$.
Let us assume the length of the sides of the triangle are $a,b$ and $c$,
So the perimeter of the triangle will be,
$Perimeter=a+b+c$
The sines of the angles of the triangle will be $\sin A,\sin B$ and $\sin C$. So its arithmetic mean will be,
$AM=\dfrac{\sin A+\sin B+\sin C}{3}$
Now we will write the equation for the given condition,
$\begin{align}
  & a+b+c=6\left( \dfrac{\sin A+\sin B+\sin C}{3} \right) \\
 & a+b+c=2\left( \sin A+\sin B+\sin C \right)..............(i)
\end{align}$


Now from Law of sines,
$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k$
We can determine,
$a=k\sin A$……. (ii)
$b=k\sin B$……. (iii)
$c=k\sin C$……(iv)
We will substitute the equations (ii), (iii) and (iv) in equation (i).
$\begin{align}
  & k\sin A+k\sin B+k\sin C=2\left( \sin A+\sin B+\sin C \right). \\
 & k\left( \sin A+\sin B+\sin C \right)=2\left( \sin A+\sin B+\sin C \right) \\
 & k=2
\end{align}$
We have to find the value of angle $A$, so we will substitute the value of $k=2$ in equation (ii) and $a=1$ as given.
$\begin{align}
  & 1=2\sin A \\
 & \sin A=\dfrac{1}{2} \\
 & \sin A=\sin \dfrac{\pi }{6} \\
 & A=\dfrac{\pi }{6}
\end{align}$

The angle $A$ of $\vartriangle ABC$ in which perimeter of is $6$ times the arithmetic mean of the sines of its angles and the side $a$ of triangle is $1$, then the angle $A$ is $A=\dfrac{\pi }{6}$. Hence the correct option is (A).

Note: The arithmetic mean can be defined as the ratio of sum of all the observations to the number of all the observations and perimeter can be defined as the sum of all the sides of the polygon.