If alpha, beta and gamma are the lengths of t | vedclass

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(A) We know that the area of a triangle is given by $\Delta = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{altitude}$.Thus,$\Delta = \frac{1}{2} a \alp

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$\sin^2 A + \sin^2 B + \sin^2 C$

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(A) We know that the area of a triangle is given by $\Delta = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{altitude}$.Thus,$\Delta = \frac{1}{2} a \alpha = \frac{1}{2} b \beta = \frac{1}{2} c \gamma$.This implies $\alpha = \frac{2 \Delta}{a}$,$\beta = \frac{2 \Delta}{b}$,and $\gamma = \frac{2 \Delta}{c}$.Substituting these into the expression:$\frac{\Delta^2}{R^2}\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}ight) = \frac{\Delta^2}{R^2}\left(\frac{a^2}{4 \Delta^2}+\frac{b^2}{4 \Delta^2}+\frac{c^2}{4 \Delta^2}ight)$$= \frac{\Delta^2}{R^2} \cdot \frac{1}{4 \Delta^2} (a^2 + b^2 + c^2) = \frac{a^2 + b^2 + c^2}{4 R^2}$.Using the sine rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.Therefore,$\frac{(2R \sin A)^2 + (2R \sin B)^2 + (2R \sin C)^2}{4 R^2} = \frac{4 R^2 (\sin^2 A + \sin^2 B + \sin^2 C)}{4 R^2} = \sin^2 A + \sin^2 B + \sin^2 C$.

If $\alpha, \beta$ and $\gamma$ are the lengths of the altitudes of a $\triangle ABC$ with area $\Delta$,then $\frac{\Delta^2}{R^2}\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}\right)$ is equal to

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