"If the angle made by the tangent at the point (x0, y0) on the curve $x = 12(t + \sin t\cos t)$, $y = 12{(1 + \sin t)^2}$, $0 0 is equal to:"

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$\because$ $${{dy} \over {dx}} = {{24(1 + \sin t)\cos t} \over {12(1 + \cos 2t)}} = {{1 + \sin t} \over {\cos t}} = \tan \left( {{\pi \over 4} + {t \over 2}} \right)$$

$\because$ $${{dy} \over {d{x_{({x_0},{y_0})}}}} = \sqrt 3 = \tan \left( {{\pi \over 4} + {t \over 2}} \right)$$

$\Rightarrow t = {\pi \over 6}$

So, $${y_{0\left( {at\,t = {\pi \over 6}} \right)}} = 12{\left( {1 + \sin {\pi \over 6}} \right)^2} = 27$$

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If the angle made by the tangent at the point (x₀, y₀) on the curve $x = 12(t + \sin t\cos t)$, $y = 12{(1 + \sin t)^2}$, $0 < t < {\pi \over 2}$, with the positive x-axis is ${\pi \over 3}$, then y₀ is equal to:

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If the angle made by the tangent at the point (x0, y0) on the curve $x = 12(t + \sin t\cos t)$, $y = 12{(1 + \sin t)^2}$, $0 < t < {\pi \over 2}$, with the positive x-axis is ${\pi \over 3}$, then y0 is equal to:

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If the angle made by the tangent at the point (x₀, y₀) on the curve $x = 12(t + \sin t\cos t)$, $y = 12{(1 + \sin t)^2}$, $0 < t < {\pi \over 2}$, with the positive x-axis is ${\pi \over 3}$, then y₀ is equal to: