
In a triangle \[ABC\], if \[B = 3C\], then the values of\[\sqrt {(\dfrac{{b + c}}{{4c}})} \]and \[\left( {\dfrac{{b - c}}{{2c}}} \right)\] are
A. \[sinC,sin\dfrac{A}{2}\]
B. \[cosC,sin\dfrac{A}{2}\]
C. \[sinC,cos\dfrac{A}{2}\]
D. None of these
Answer
161.1k+ views
Hint: There doesn't seem to be a direct solution given the constraints. However, if we replace each term with its trigonometric equivalent and simplify, then it becomes: \[\left( {\dfrac{{b + c}}{{4c}}} \right) = \sqrt {sin{A^2} + cosC{\rm{ }}sin{B^2}} \]
and \[\left( {\dfrac{{b - c}}{{2c}}} \right) = \left( {1/{\rm{ }}cosC} \right) + \left( {1/{\rm{ }}sinB2} \right)\].
Any constraints you are provided in the problem are represented by the constraint equation. The problem may or may not always have a constraint. This could mean that one variable in the objective equation already exists. Only one of these curves is the one we refer to as straight which has the shortest length out of all of these curves. The triangle inequality, which asserts that the total of any two sides of a triangle is greater than the length of the third side, is at the core of this condition.
A parameter or variable is bound by constraints, which set upper and lower bounds. The expression of variable constraints can take the form of absolute values, functions of parameters, or variables in the initial conditions. The variable declarations section also contains the beginning condition and a variable constraint.
Complete step by step solution: The given equation is,
\[\sqrt {\dfrac{{b + c}}{{4c}}} \]
\[ = \sqrt {\dfrac{{sin3C + \sin C}}{{4\sin C}}} \]
We have been provided in the question that in a triangle ABC,
\[B = {\rm{sin}}3C\]
Now, we have to take sin on either side of the above equation, we get
\[\sin B = {\rm{sin}}3C\]
By sin formula, we get
\[\sin B = 3\sin C - 4\sin 3\;C\]
Now, we have to replace the value as per the question in the given formula, we get
\[ = > \sqrt {\dfrac{{3sin2C + \cos C}}{{4\sin C}}} \]
The equation can also be written as,
\[ = > \sqrt {\dfrac{{2sin2C\cos C}}{{4\sin C}}} \]
Then it becomes,
\[\cos C\]
\[\dfrac{{b - c}}{{2c}} = \dfrac{{\sin 3C - \sin C}}{{2\sin C}}\]
\[ = \dfrac{{2\cos 2C\sin C}}{{2\sin C}}\]
By solving this, the equation becomes
\[\cos 2C\]
So, \[\sin \dfrac{A}{2}\]
So, Option ‘B’ is correct
Note: The notes for the problems are as follows:
-In a triangle \[ABC\], if \[B = 3C\], then the values of \[\sqrt {\left( {\dfrac{{b + c}}{{4c}}} \right)} \] and \[\left( {\dfrac{{b - c}}{{2c}}} \right)\] are \[cosC,sin{A^2}\].
-The value of √ is \[1.41\] and \[sin{A^2}\] is \[ - 1\].
The triangle inequality can be used to calculate the best upper estimate of the size of the sum of two numbers in terms of the sizes of the separate numbers in mathematical analysis. If and only if the vector y is a nonnegative scalar of the vector \[x\].
and \[\left( {\dfrac{{b - c}}{{2c}}} \right) = \left( {1/{\rm{ }}cosC} \right) + \left( {1/{\rm{ }}sinB2} \right)\].
Any constraints you are provided in the problem are represented by the constraint equation. The problem may or may not always have a constraint. This could mean that one variable in the objective equation already exists. Only one of these curves is the one we refer to as straight which has the shortest length out of all of these curves. The triangle inequality, which asserts that the total of any two sides of a triangle is greater than the length of the third side, is at the core of this condition.
A parameter or variable is bound by constraints, which set upper and lower bounds. The expression of variable constraints can take the form of absolute values, functions of parameters, or variables in the initial conditions. The variable declarations section also contains the beginning condition and a variable constraint.
Complete step by step solution: The given equation is,
\[\sqrt {\dfrac{{b + c}}{{4c}}} \]
\[ = \sqrt {\dfrac{{sin3C + \sin C}}{{4\sin C}}} \]
We have been provided in the question that in a triangle ABC,
\[B = {\rm{sin}}3C\]
Now, we have to take sin on either side of the above equation, we get
\[\sin B = {\rm{sin}}3C\]
By sin formula, we get
\[\sin B = 3\sin C - 4\sin 3\;C\]
Now, we have to replace the value as per the question in the given formula, we get
\[ = > \sqrt {\dfrac{{3sin2C + \cos C}}{{4\sin C}}} \]
The equation can also be written as,
\[ = > \sqrt {\dfrac{{2sin2C\cos C}}{{4\sin C}}} \]
Then it becomes,
\[\cos C\]
\[\dfrac{{b - c}}{{2c}} = \dfrac{{\sin 3C - \sin C}}{{2\sin C}}\]
\[ = \dfrac{{2\cos 2C\sin C}}{{2\sin C}}\]
By solving this, the equation becomes
\[\cos 2C\]
So, \[\sin \dfrac{A}{2}\]
So, Option ‘B’ is correct
Note: The notes for the problems are as follows:
-In a triangle \[ABC\], if \[B = 3C\], then the values of \[\sqrt {\left( {\dfrac{{b + c}}{{4c}}} \right)} \] and \[\left( {\dfrac{{b - c}}{{2c}}} \right)\] are \[cosC,sin{A^2}\].
-The value of √ is \[1.41\] and \[sin{A^2}\] is \[ - 1\].
The triangle inequality can be used to calculate the best upper estimate of the size of the sum of two numbers in terms of the sizes of the separate numbers in mathematical analysis. If and only if the vector y is a nonnegative scalar of the vector \[x\].
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

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

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

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

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

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 In Hindi Chapter 1 Sets

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations
