If Aleft(frac{pi}{3}right) and Bleft(frac{pi} | vedclass

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Question

(A) The parametric equations of the given circle are $x = 12 \cos 	heta$ and $y = 12 \sin 	heta$. For point $A$ with parameter $	heta = \frac{\pi}{

Vedclass · Accepted Answer

$6(\sqrt{6}-\sqrt{2})$

Vedclass · Answer

(A) The parametric equations of the given circle are $x = 12 \cos 	heta$ and $y = 12 \sin 	heta$. For point $A$ with parameter $	heta = \frac{\pi}{3}$: $x_A = 12 \cos \frac{\pi}{3} = 12 	imes \frac{1}{2} = 6$ $y_A = 12 \sin \frac{\pi}{3} = 12 	imes \frac{\sqrt{3}}{2} = 6\sqrt{3}$ So,$A = (6, 6\sqrt{3})$. For point $B$ with parameter $	heta = \frac{\pi}{6}$: $x_B = 12 \cos \frac{\pi}{6} = 12 	imes \frac{\sqrt{3}}{2} = 6\sqrt{3}$ $y_B = 12 \sin \frac{\pi}{6} = 12 	imes \frac{1}{2} = 6$ So,$B = (6\sqrt{3}, 6)$. The length of the chord $AB$ is given by the distance formula: $AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$ $AB = \sqrt{(6\sqrt{3} - 6)^2 + (6 - 6\sqrt{3})^2}$ $AB = \sqrt{2 	imes (6\sqrt{3} - 6)^2}$ $AB = \sqrt{2} 	imes |6\sqrt{3} - 6|$ $AB = 6\sqrt{2}(\sqrt{3} - 1)$ $AB = 6(\sqrt{6} - \sqrt{2})$.

If $A\left(\frac{\pi}{3}\right)$ and $B\left(\frac{\pi}{6}\right)$ are points on a circle represented in parametric form with center $(0,0)$ and radius $12$,then the length of the chord $AB$ is:

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