Question

The sides of a triangle inscribed in a given circle subtend angles $\alpha ,\beta ,\gamma$ at the centre. 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?

• A. $-\frac{\sqrt{3}}{2}$
• B. $\frac{\sqrt{3}}{2}$
• C. $-\frac{2}{\sqrt{3}}$
• D. $\sqrt{2}$

Answer 1

The correct option is A $-\frac{\sqrt{3}}{2}$

To find AM of $\cos (\alpha +\frac{\pi }{2}),\cos (\beta +\frac{\pi }{2}),\cos (\gamma +\frac{\pi }{2})$

We know,

$AM\ge GM$

Hence minimum value of AM is its GM.

This implies,

$\cos (\alpha +\frac{\pi }{2})=\cos (\beta +\frac{\pi }{2})=\cos (\gamma +\frac{\pi }{2})$

$-\sin \alpha =-\sin \beta =-\sin \gamma$

$\alpha =\beta =\gamma$

At center,

$\alpha +\beta +\gamma =2\pi$

$\alpha +\alpha +\alpha ={360}^0$

$\alpha =\frac{2\pi }{3}=\beta =\gamma$

$AM=\frac{\cos (\alpha +\frac{\pi }{2})+\cos (\beta +\frac{\pi }{2})+\cos (\gamma +\frac{\pi }{2})}{3}$

Substituting values,

$AM=\frac{\cos (\frac{2\pi }{3}+\frac{\pi }{2})+\cos (\frac{2\pi }{3}+\frac{\pi }{2})+\cos (\frac{2\pi }{3}+\frac{\pi }{2})}{3}$

$=3\times \frac{1}{3}\times \cos \left({210}^0\right)=-\frac{\sqrt{3}}{2}$

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