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(B) In $	riangle CPQ$,$CP = CQ = \alpha$ and $\angle PCQ = 2	heta$. The length of the base $PQ = 2\alpha \sin 	heta$.The area of $	riangle CPQ$ is

Vedclass · Accepted Answer

$\sin 	heta = \frac{\sqrt{5} - 1}{2}$

Vedclass · Answer

(B) In $	riangle CPQ$,$CP = CQ = \alpha$ and $\angle PCQ = 2	heta$. The length of the base $PQ = 2\alpha \sin 	heta$.The area of $	riangle CPQ$ is $\Delta = \frac{1}{2} \alpha^2 \sin 2	heta$.The semi-perimeter $s = \frac{CP + CQ + PQ}{2} = \frac{\alpha + \alpha + 2\alpha \sin 	heta}{2} = \alpha(1 + \sin 	heta)$.The inradius $r$ is given by $r = \frac{\Delta}{s} = \frac{\frac{1}{2} \alpha^2 \sin 2	heta}{\alpha(1 + \sin 	heta)} = \frac{\alpha}{2} \cdot \frac{2 \sin 	heta \cos 	heta}{1 + \sin 	heta} = \alpha \cdot \frac{\sin 	heta \cos 	heta}{1 + \sin 	heta}$.To maximize $r$,we maximize $f(	heta) = \frac{\sin 	heta \cos 	heta}{1 + \sin 	heta}$.Let $x = \sin 	heta$. Then $\cos 	heta = \sqrt{1 - x^2}$. So $f(x) = \frac{x \sqrt{1 - x^2}}{1 + x} = x \sqrt{\frac{1 - x}{1 + x}}$.Squaring $f(x)$,we maximize $g(x) = x^2 \frac{1 - x}{1 + x} = \frac{x^2 - x^3}{1 + x}$.Using $g'(x) = 0$: $\frac{(2x - 3x^2)(1 + x) - (x^2 - x^3)(1)}{(1 + x)^2} = 0$.$(2x + 2x^2 - 3x^2 - 3x^3) - (x^2 - x^3) = 0 \implies -2x^3 - 2x^2 + 2x = 0$.$-2x(x^2 + x - 1) = 0$. Since $x = \sin 	heta > 0$,we have $x^2 + x - 1 = 0$.Solving for $x$,$x = \frac{-1 \pm \sqrt{1 + 4}}{2}$. Since $x > 0$,$x = \sin 	heta = \frac{\sqrt{5} - 1}{2}$.

$P$ and $Q$ are two points on a circle with center $C$ and radius $\alpha$. The angle $\angle PCQ = 2\theta$. The radius $r$ of the circle inscribed in the triangle $CPQ$ is maximum when:

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