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

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 \[x\].