The sides of a triangle inscribed in a given | vedclass

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(B) Let the triangle be $ABC$. Since the angle made by a chord at the center of a circle is twice the angle made by the chord at the circumference,we

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$-\frac{\sqrt{3}}{2}$

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(B) Let the triangle be $ABC$. Since the angle made by a chord at the center of a circle is twice the angle made by the chord at the circumference,we have $\angle A = \frac{\alpha}{2}$,$\angle B = \frac{\beta}{2}$,$\angle C = \frac{\gamma}{2}$. Since $A+B+C = \pi$,we have $\alpha + \beta + \gamma = 2\pi$. The $A.M.$ of the given terms is $\frac{1}{3} [\cos (\alpha + \frac{\pi}{2}) + \cos (\beta + \frac{\pi}{2}) + \cos (\gamma + \frac{\pi}{2})]$. Using $\cos(	heta + \frac{\pi}{2}) = -\sin 	heta$,this becomes $-\frac{1}{3} [\sin \alpha + \sin \beta + \sin \gamma]$. Using $\sin \alpha + \sin \beta + \sin \gamma = 4 \sin \frac{\alpha}{2} \sin \frac{\beta}{2} \sin \frac{\gamma}{2} = 4 \sin A \sin B \sin C$,the $A.M.$ is $-\frac{4}{3} \sin A \sin B \sin C$. For a fixed sum $A+B+C = \pi$,the product $\sin A \sin B \sin C$ is maximum when $A=B=C = \frac{\pi}{3}$. Thus,the minimum value is $-\frac{4}{3} (\sin \frac{\pi}{3})^3 = -\frac{4}{3} (\frac{\sqrt{3}}{2})^3 = -\frac{4}{3} \cdot \frac{3\sqrt{3}}{8} = -\frac{\sqrt{3}}{2}$.

The sides of a triangle inscribed in a given circle subtend angles $\alpha, \beta, \gamma$ at the center. The minimum value of the $A.M.$ of $\cos (\alpha + \frac{\pi}{2})$,$\cos (\beta + \frac{\pi}{2})$ and $\cos (\gamma + \frac{\pi}{2})$ is equal to

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