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(A) Let $R$ be the radius of the semi-circle. Let $\angle AOY = 	heta$ and $\angle BOY = 	heta$. Since $AB \parallel XY$,the points $A$ and $B$ are

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$\cos^{-1}\left(\frac{\sqrt{5}-1}{2}\right)$

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(A) Let $R$ be the radius of the semi-circle. Let $\angle AOY = 	heta$ and $\angle BOY = 	heta$. Since $AB \parallel XY$,the points $A$ and $B$ are symmetric with respect to the perpendicular bisector of $XY$. The coordinates can be represented as $A = (R \cos 	heta, R \sin 	heta)$ and $B = (-R \cos 	heta, R \sin 	heta)$.The sides of $	riangle AOB$ are $OA = R$,$OB = R$,and $AB = 2R \cos 	heta$.The semi-perimeter $s = \frac{R + R + 2R \cos 	heta}{2} = R(1 + \cos 	heta)$.The area of $	riangle AOB$ is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2R \cos 	heta) 	imes (R \sin 	heta) = R^2 \sin 	heta \cos 	heta$.The inradius $r = \frac{	ext{Area}}{s} = \frac{R^2 \sin 	heta \cos 	heta}{R(1 + \cos 	heta)} = R \frac{\sin 	heta \cos 	heta}{1 + \cos 	heta}$.To maximize $r$,we differentiate with respect to $	heta$ and set to $0$:$\frac{dr}{d	heta} = R \frac{(1 + \cos 	heta)(\cos^2 	heta - \sin^2 	heta) - (\sin 	heta \cos 	heta)(-\sin 	heta)}{(1 + \cos 	heta)^2} = 0$.This simplifies to $(1 + \cos 	heta)(2 \cos^2 	heta - 1) + \sin^2 	heta \cos 	heta = 0$.Using $\sin^2 	heta = 1 - \cos^2 	heta$,we get $(1 + \cos 	heta)(2 \cos^2 	heta - 1) + (1 - \cos^2 	heta) \cos 	heta = 0$.$2 \cos^2 	heta - 1 + 2 \cos^3 	heta - \cos 	heta + \cos 	heta - \cos^3 	heta = 0$.$\cos^3 	heta + 2 \cos^2 	heta - 1 = 0$.Factoring gives $(\cos 	heta + 1)(\cos^2 	heta + \cos 	heta - 1) = 0$.Since $\cos 	heta 
eq -1$,we have $\cos^2 	heta + \cos 	heta - 1 = 0$.Solving for $\cos 	heta$,we get $\cos 	heta = \frac{-1 \pm \sqrt{1 + 4}}{2}$. Since $	heta$ is acute,$\cos 	heta = \frac{\sqrt{5}-1}{2}$.Thus,$\angle BOY = 	heta = \cos^{-1}\left(\frac{\sqrt{5}-1}{2}ight)$.

Let $XY$ be the diameter of a semi-circle with center $O$. Let $A$ be a variable point on the semi-circle and $B$ another point on the semi-circle such that $AB$ is parallel to $XY$. The value of $\angle BOY$ for which the inradius of $\triangle AOB$ is maximum,is

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