Question

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

• A. ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C$
• B. ${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C$
• C. ${\tan }^{2}A+{\tan }^{2}B+{\tan }^{2}C$
• D. ${\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C$

Answer 1

The correct option is A ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C$

$\frac{{\Delta }^2}{R^2}(\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2})=?.....\left(1\right)$

$\Delta =\frac{1}{2}a\alpha =\frac{1}{2}b\beta =\frac{1}{2}c\gamma$

(Using $\Delta =\frac{1}{2}\times base\times height)$

$\alpha =\frac{2\Delta }{a},\beta =\frac{2\Delta }{b},\gamma =\frac{2\Delta }{c}$

Substituting in equating (1), we get

$\Rightarrow \frac{{\Delta }^2}{R^2}(\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2})=\frac{4}{R^2}(a^2+b^2+c^2)$

$=\frac{4R^2({\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C)}{4R^2}$

$={\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C$