Question

$\Delta$ ABC is not right-angled and is inscribed in a fixed circle. If $a,A,b,B$ be slightly varied keeping $c,C$ fixed, then $\frac{da}{cosA}+\frac{db}{cosB}=?$

• A. $2R$
• B. $\pi$
• C. $0$
• D. none of these

Answer 1

Answer: (C)

$0$

Given that side c and angle C are constant.
$\therefore \frac{c}{\sin C}=K\Rightarrow \frac{a}{\sin A}=\frac{b}{\sin B}=K\Rightarrow a=K\sin Aandb=K\sin B\Rightarrow \frac{da}{dA}=K\cos Aand\frac{db}{dB}=K\cos B\Rightarrow \frac{da}{\cos A}+\frac{db}{\cos B}=K(dA+dB)$.........(1)
Also for the triangle ABC, $A+B+C=\pi$
$\Rightarrow A+B=\pi -C$
$\Rightarrow dA+dB=0$ (since C is fixed)..........(2)
From (1) and (2), $\frac{da}{\cos A}+\frac{db}{\cos B}=0$