Acute angle between the lines represented by the $({x^2} + {y^2})\sqrt 3 = 4xy$ is
A. $\dfrac{\pi }{6}$
B. $\dfrac{\pi }{4}$
C. $\dfrac{\pi }{3}$
D. None of these

Hint: We are given two lines in this case, and we must determine the acute angle between them. They are in slope intercept form as\[y = mx + c\] where m represents the slope of the line, we shall first determine its slope. The angle between the two lines will then be determined using the formula \[tan\theta {\rm{ }} = \;\frac{{{m_2} - {m_1}}}{{1 + {m_1} \cdot {m_2}}}\] after we have determined the slope of both lines. By resolving this, we shall obtain the result.

Formula Used:The tangent function's formula can be used to determine the acute angle between the lines.\[tan\theta {\rm{ }} = \;\frac{{{m_1} - {m_2}}}{{1 + {m_1} \cdot {m_2}}}\]

Complete step by step solution: Here, it is given the equation that
\[\left( {{x^2} + {y^2}} \right)\sqrt 3 = 4xy\]
And we should determine the acute angle between the lines represented by \[\left( {{x^2} + {y^2}} \right)\sqrt 3 = 4xy\]
Now, let us write the two lines that represent the given equation, we have
\[y - {m_1}x = 0\]
AND
\[y - {m_2}x = 0\]
Now, we have to write the combined equation of the above two lines, we get
\[\left( {y - {m_1}x} \right)\left( {y - {m_2}x} \right) = 0\]
Now, we have to multiply both the equations, we have
\[{y^2} - \left( {{m_1} + {m_2}} \right)\left( {xy} \right) + {m_1}{m_2}{x^2} = 0\]------- (1)
Now, we have to consider the given equation
\[\left( {{x^2} + {y^2}} \right)\sqrt 3 = 4xy\]
Let us write the above equation in quadratic form, we get
\[\left( {{x^2} + {y^2}} \right) - \frac{4}{{\sqrt 3 }}xy = 0\]
Now, we have to compare the above equation with the equation (1), we have
\[{m_1} + {m_2} = \frac{4}{{\sqrt 3 }}\]
Since, we already know the condition that,
\[{m_1}{m_2} = 1\]
Therefore, we have
\[{\left( {{m_1} - {m_2}} \right)^2} = {\left( {{m_1} + {m_2}} \right)^2} - 4{m_1}{m_2}\]
Now, we have to substitute the values of each term, we have
\[ \Rightarrow {\left( {{m_1} - {m_2}} \right)^2} = \frac{{16}}{3} - 4\]
On simplifying the above equation, we get
\[ \Rightarrow {\left( {{m_1} - {m_2}} \right)^2} = \frac{4}{3}\]
Now, we have to remove the square by taking modulus, we get
\[\left| {{m_1} - {m_2}} \right| = \frac{2}{{\sqrt 3 }}\]
Now, we have an acute angle between these two lines, we have
\[\tan \alpha = \left| {\frac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|\]
Now, we have to substitute the values of each term, we have
\[\tan \alpha = \left| {\frac{{\frac{2}{{\sqrt 3 }}}}{{1 + 1}}} \right| = \frac{1}{{\sqrt 3 }}\]
By trigonometry identity, we can write the value as below
\[\tan \alpha = \tan \left( {\frac{\pi }{6}} \right)\]
On cancelling the similar terms from the above equation, we get
\[ \Rightarrow \alpha = \frac{\pi }{6}\]
Therefore, acute angle between the lines represented by \[\left( {{x^2} + {y^2}} \right)\sqrt 3 = 4xy\] is \[\frac{\pi }{6}\]
Hence, the option A is correct

Note: Students often tend to make mistake in these types of problems because it includes formula to remember. Here, one should keep in mind that calculation part and applications of formulas are very important in order to get the desired answer. Make sure you won't commit this error since sometimes even examiners are quite savvy and will provide you options where you can observe that obtuse angle and mark it as the correct answer.