The circle x=5 cos theta, y=5 sin theta is bo | vedclass

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(A) The coordinate of $P$ is $\left(5 \cos \frac{2 \pi}{3}, 5 \sin \frac{2 \pi}{3}\right) = \left(-\frac{5}{2}, \frac{5 \sqrt{3}}{2}\right)$.The slope

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$\frac{62-24 \sqrt{3}}{\sqrt{3}}$

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(A) The coordinate of $P$ is $\left(5 \cos \frac{2 \pi}{3}, 5 \sin \frac{2 \pi}{3}ight) = \left(-\frac{5}{2}, \frac{5 \sqrt{3}}{2}ight)$.The slope of the tangent at $	heta = \frac{2 \pi}{3}$ is $m = -\cot \left(\frac{2 \pi}{3}ight) = \frac{1}{\sqrt{3}}$.The equation of the tangent line is $y - \frac{5 \sqrt{3}}{2} = \frac{1}{\sqrt{3}}\left(x + \frac{5}{2}ight)$,which simplifies to $x - \sqrt{3}y + 10 = 0$.To find the intersection point $Q$ with the line $y = 6$,we substitute $y = 6$ into the tangent equation: $x - 6\sqrt{3} + 10 = 0 \Rightarrow x = 6\sqrt{3} - 10$. Thus,$Q = (6\sqrt{3} - 10, 6)$.To find the intersection point $R$ with the line $x = -6$,we substitute $x = -6$ into the tangent equation: $-6 - \sqrt{3}y + 10 = 0$ $\Rightarrow \sqrt{3}y = 4$ $\Rightarrow y = \frac{4}{\sqrt{3}}$. Thus,$R = (-6, \frac{4}{\sqrt{3}})$.The third vertex of the triangle is $S = (-6, 6)$.The area of $	riangle RSQ$ is given by $\frac{1}{2} |x_R(y_S - y_Q) + x_S(y_Q - y_R) + x_Q(y_R - y_S)|$.Area $= \frac{1}{2} |(-6)(6 - 6) + (-6)(6 - \frac{4}{\sqrt{3}}) + (6\sqrt{3} - 10)(\frac{4}{\sqrt{3}} - 6)|$.Area $= \frac{1}{2} |0 - 36 + \frac{24}{\sqrt{3}} + (24 - 36\sqrt{3} - \frac{40}{\sqrt{3}} + 60)| = \frac{1}{2} |48 - 36\sqrt{3} - \frac{16}{\sqrt{3}}| = |24 - 18\sqrt{3} - \frac{8}{\sqrt{3}}| = |\frac{24\sqrt{3} - 54 - 8}{\sqrt{3}}| = \frac{62 - 24\sqrt{3}}{\sqrt{3}}$.

The circle $x=5 \cos \theta, y=5 \sin \theta$ is bounded by the rectangle formed by the lines $x \pm 6=0$ and $y \pm 6=0$. The area of the triangle that lies inside the rectangle,which is formed by the tangent at $P\left(\frac{2 \pi}{3}\right)$ to the circle with two of the above given lines,is

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