Find the angle between the lines \[x\cos30^{\circ} + y\sin30^{\circ} = 3\] and \[x\cos60^{\circ} + y\sin60^{\circ} = 5\].
A. \[90^{\circ} \]
B. \[30^{\circ} \]
C. \[60^{\circ} \]
D. None of these

Hint First rewrite the given equations in the form \[y = mx + c\]. Then substitute the values of the trigonometric angles. After that, determine the slope of each line. In the end, use the formula \[\theta = \tan^{ - 1}\left| {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|\] to get the angle between the two lines.

Formula used
The angle between the two lines with slope \[{m_1}\] and \[{m_2}\] is: \[\theta = \tan^{ - 1}\left| {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|\]
\[\sin30^{\circ} = \dfrac{1}{2}\]
\[\cot30^{\circ} = \sqrt{3} \]
\[\sin60^{\circ} = \dfrac{{\sqrt 3 }}{2}\]
\[\cot60^{\circ} = \dfrac{1}{{\sqrt 3 }}\]

Complete step by step solution:
The given equations of the lines are \[x\cos30^{\circ} + y\sin30^{\circ} = 3\] and \[x\cos60^{\circ} + y\sin60^{\circ} = 5\].
Let’s simplify the equations in the point-slope form.
\[x\cos30^{\circ} + y\sin30^{\circ} = 3\]
\[ \Rightarrow \]\[y\sin30^{\circ} = 3 – x\cos30^{\circ} \]
Divide both sides by \[\sin30^{\circ} \]
\[y = \dfrac{3}{{\sin30^{\circ} }} - \dfrac{{x\cos30^{\circ} }}{{\sin30^{\circ} }}\]
\[ \Rightarrow \]\[y = \dfrac{3}{{\sin30^{\circ} }} – x\cot30^{\circ} \] [ Since \[\dfrac{{\cos x}}{{\sin x}} = \cot x\]]
\[ \Rightarrow \]\[y = \dfrac{3}{{\dfrac{1}{2}}} - x\left( {\sqrt 3 } \right)\] [ Since \[\cot30^{\circ} = \sqrt 3 \] and \[\sin30^{\circ} = \dfrac{1}{2}\]]
\[ \Rightarrow \]\[y = - \sqrt 3 x + 6\]
Therefore, the slope of the line is \[{m_1} = - \sqrt 3 \].
\[x\cos60^{\circ} + y\sin60^{\circ} = 5\]
\[ \Rightarrow \]\[y\sin60^{\circ} = 5 – x\cos60^{\circ} \]
Divide both sides by \[\sin60^{\circ} \]
\[y = \dfrac{5}{{\sin60^{\circ} }} - \dfrac{{x\cos60^{\circ} }}{{\sin60^{\circ} }}\]
\[ \Rightarrow \]\[y = \dfrac{5}{{\sin60^{\circ} }} – x\cot60^{\circ} \] [ Since \[\dfrac{{\cos x}}{{\sin x}} = \cot x\]]
\[ \Rightarrow \]\[y = \dfrac{5}{{\dfrac{{\sqrt 3 }}{2}}} - x\left( {\dfrac{1}{{\sqrt 3 }}} \right)\] [ Since \[\cot60^{\circ} = \dfrac{1}{{\sqrt 3 }}\] and \[\sin60^{\circ} = \dfrac{{\sqrt 3 }}{2}\]]
\[ \Rightarrow \]\[y = - \dfrac{1}{{\sqrt 3 }}x + \dfrac{{10}}{{\sqrt 3 }}\]
Therefore, the slope of the line is \[{m_2} = - \dfrac{1}{{\sqrt 3 }}\].
Now apply the formula of the angle between the two lines \[\theta = \tan^{ - 1}\left| {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|\].
Substitute the values of the slopes in the formula.
\[\theta = \tan^{ - 1}\left| {\dfrac{{ - \dfrac{1}{{\sqrt 3 }} - \left( { - \sqrt 3 } \right)}}{{1 + \left( { - \sqrt 3 } \right)\left( { - \dfrac{1}{{\sqrt 3 }}} \right)}}} \right|\]
Simplify the above equation.
\[\theta = \tan^{ - 1}\left| {\dfrac{{ - \dfrac{1}{{\sqrt 3 }} + \sqrt 3 }}{{1 + 1}}} \right|\]
\[ \Rightarrow \]\[\theta = \tan^{ - 1}\left| {\dfrac{{\dfrac{{ - 1 + 3}}{{\sqrt 3 }}}}{2}} \right|\]
\[ \Rightarrow \]\[\theta = \tan^{ - 1}\left| {\dfrac{{\dfrac{2}{{\sqrt 3 }}}}{2}} \right|\]
\[ \Rightarrow \]\[\theta = \tan^{ - 1}\left| {\dfrac{1}{{\sqrt 3 }}} \right|\]
\[ \Rightarrow \]\[\theta = \tan^{ - 1}\left( {tan3{0^o}} \right)\] [Since \[tan30^{\circ} = \dfrac{1}{{\sqrt 3 }}\]]
\[ \Rightarrow \]\[\theta = 30^{\circ} \][Since \[\tan^{ - 1}\left( {tanA} \right) = A\]]
Hence the correct option is B.

Note: The angle between the two lines can be found by calculating the slope of each line and then using them in the formula to determine the angle between two lines when the slope of each line is known.