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
$\frac{a}{2\,sin\,A}=\frac{b}{2\,sin\,B}=\frac{c}{2\,sin\,C}=R$
$\therefore da=2\,R\,cos\,A\,dA$
$db=2\,R\,cos\,B\,dB$
$dc=2R\,cos\,C\,dC$
$\therefore \frac{da}{cos\,A}+\frac{db}{cos\,B}+\frac{dc}{cos\,C}$
$=2\,R\left[dA+dB+dC\right]$
Also $A+B+C=\pi$ ( = constant)
$\therefore dA + dB + dC = 0$
$\therefore \frac{da}{coa\,A}+\frac{db}{cos\,B}+\frac{dc}{cos\,C}=0$