Explanation
x=a\left( \theta +\sin { \theta } \right)
y=a\left( 1+\cos { \theta } \right)
If tangent to curve at \theta = { \pi }/{ 3 }, makes angle \alpha with x-axis,
\Rightarrow slope of tangent = \tan { \alpha }
\Rightarrow \left. \begin{matrix} \tan { \alpha } ={ { \dfrac { { dy }/{ { d\theta } } }{ { dx }/{ { d\theta } } } } }{ { } } \\ \end{matrix} \right| _{ \theta ={ \pi }/{ 3 } }
=\dfrac {-a\sin \theta}{ a\left( 1+\cos { \theta } \right) }
\left. \begin{matrix}\dfrac{-\sin \theta} { 1+\cos { \theta } } \\ \end{matrix}\right|_{ \theta ={ \pi }/{ 3 } }
=-\dfrac{\sqrt{3}/2} { 1+{ 1 }/{ 2 } }
\tan { \alpha } \ =-\dfrac{1}{\sqrt { 3 }}
Ans. \alpha \ =\ { 5{ \pi } }/{ 6 }
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