
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
161.4k+ views
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.
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.
Recently Updated Pages
If there are 25 railway stations on a railway line class 11 maths JEE_Main

Minimum area of the circle which touches the parabolas class 11 maths JEE_Main

Which of the following is the empty set A x x is a class 11 maths JEE_Main

The number of ways of selecting two squares on chessboard class 11 maths JEE_Main

Find the points common to the hyperbola 25x2 9y2 2-class-11-maths-JEE_Main

A box contains 6 balls which may be all of different class 11 maths JEE_Main

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

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

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

NIT Cutoff Percentile for 2025

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

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025
