If the angle made by the tangent at the point $\left(x_0,y_0\right)$ on the curve $x=12\left(t+\sin t\cos t\right)$,
$y=12{\left(1+\sin t\right)}^2,0<t<\frac{\pi }{2}$, with the positive $x$-axis is $\frac{\pi }{3}$, then $y_0$ is equal to

Given, 

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

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

Now $\frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}$

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

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

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

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

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

$\frac{\pi }{3}=\frac{t}{2}+\frac{\pi }{4} \Rightarrow \frac{t}{2}=\frac{\pi }{12} \Rightarrow t=\frac{\pi }{6}$

Now putting $t=\frac{\pi }{6}$ in $y=12{\left(1+\sin t\right)}^2$

as $\left(x_0,y_0\right)$ lie on curve

So $y_0=12{\left(1+\sin \frac{\pi }{6}\right)}^2$

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