Question

If the sides of a triangle $ABC$ vary slightly in such a way so that its circumradius remains constant, then $\frac{da}{\cos A}+\frac{da}{\cos B}+\frac{da}{\cos C}=$

• A. 6 R
• B. 2 R
• C. 0
• D. none of these

Answer 1

Answer: (C)

We know that
$\frac{a}{2sinA}=\frac{b}{2sinB}=\frac{c}{2sinC}=R$
$\therefore da=2RcosAdA$
$db=2RcosBdB$
$dc=2RcosCdC$
$\therefore \frac{da}{cosA}+\frac{db}{cosB}+\frac{dc}{cosC}$
$=2R\left[dA+dB+dC\right]$
Also $A+B+C=\pi$ ( = constant)
$\therefore dA+dB+dC=0$
$\therefore \frac{da}{coaA}+\frac{db}{cosB}+\frac{dc}{cosC}=0$