Question

lf $\alpha ,\beta ,\gamma$ are the altitudes of a $\Delta ABC$ then $\frac{{\alpha }^{-2}+{\beta }^{-2}+{\gamma }^{-2}}{\cot A+\cot B+\cot C}=$

• A. $\frac{1}{\Delta }$
• B. $\Delta$
• C. $\frac{1}{{\Delta }^2}$
• D. ${\Delta }^2$

Answer 1

Answer: (A)

$\frac{1}{\Delta }$

Let the Triangle be Equilateral with sides $a=b=c=1$

so, every altitude will divide the containing side into two equal parts

so, $Area=$ $△=\frac{1}{2}\alpha =\frac{1}{2}\beta =\frac{1}{2}\gamma$

$\alpha =\beta =\gamma =2△\Rightarrow \frac{1}{\alpha }=\frac{1}{\beta }=\frac{1}{\gamma }=\frac{1}{2△}$

Now, from above Formula

$cotA+cotB+cotC=\frac{1+1+1}{4△}=\frac{3}{4△}$

Hence, $\frac{{\alpha }^{-2}+{\beta }^{-2}+{\gamma }^{-2}}{cotA+cotB+cotC}=\frac{3\times \frac{1}{4{△}^2}}{\frac{3}{4△}}=\frac{1}{△}$

Therefore Answer is $A$