If the angle made by the tangent at the point ( x _ 0 , y _ 0 ) on...

Given,

Curve $x = 12\left(t + \sin t \cos t\right)$

$y = 12{\left(1 + \sin t\right)}^{2}$

Now $\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$

$\tan \left(\dfrac{\pi}{3}\right) = \dfrac{\dfrac{d}{dt}\left(12{\left(1 + \sin t\right)}^{2}\right)}{\dfrac{d}{dt}\left(12\left(t + \sin t \cos t\right)\right)}$

$\sqrt{3} = \dfrac{12 \times 2\left(1 + \sin t\right)\left(\cos t\right)}{12\left(1 + \cos 2t\right)}$

$\sqrt{3} = \dfrac{2\left(1 + \sin t\right) \cdot \cos t}{2\cos^{2}t}$

$\sqrt{3} = \dfrac{1 + \sin t}{\cos t} = \dfrac{{\left(cos\dfrac{t}{2} + sin\dfrac{t}{2}\right)}^{2}}{\cos^{2}\dfrac{t}{2} - \sin^{2}\dfrac{t}{2}}$

$\sqrt{3} = \tan \left(\dfrac{t}{2} + \dfrac{\pi}{4}\right)$

$\dfrac{\pi}{3} = \dfrac{t}{2} + \dfrac{\pi}{4} ⇒ \dfrac{t}{2} = \dfrac{\pi}{12} ⇒ t = \dfrac{\pi}{6}$

Now putting $t = \dfrac{\pi}{6}$ in y = 12{\left(1 +\text{ \sin t }\right)}^{2}$</p><p>as \left({x}_{0}, {y}_{0}\right)$ lie on curve

So ${y}_{0} = 12{\left(1 + sin\dfrac{\pi}{6}\right)}^{2}$

$= 12{\left(\dfrac{3}{2}\right)}^{2} = 27$