In a $∆ABC$ if sides a and b remain constant such that $a$ is the error in $C,$ then relative error in its area, is

Solution:

We have,

$S=\frac{1}{2}ab\sin C\Rightarrow \frac{dS}{dC}=\frac{1}{2}ab\cos C$

Let $∆S$ be the error in area $S.$ Then,

$\begin{array}{l}\mathrm{\Delta }S=\frac{dS}{dC}\mathrm{\Delta }C\\ \mathrm{\Delta }S=\frac{1}{2}ab\cos C\alpha [\because \mathrm{\Delta }C=\alpha ]\end{array}$

$\Rightarrow \frac{\mathrm{\Delta }S}{S}=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2 $}\right.}ab\cos C \alpha }{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2 $}\right. ab\sin C}= \alpha cotC$