In a $\triangle A B C$ if sides a and b remain constant such that a is the error in C, then relative error in its area, is

$\alpha \cot C$

We have,

S → Area of triangle

$S=\frac{1}{2} a b \sin C \Rightarrow \frac{d S}{d C}=\frac{1}{2} a b \cos C$

Let $\Delta S$ be the error in area S. Then,

$\Delta S =\frac{d S}{d C} \Delta C$          $[∵ \Delta C=\alpha]$

$\Rightarrow \Delta S =\frac{1}{2} a b ~\cos C ~\alpha$

$\Rightarrow \frac{\Delta S}{S}=\frac{\frac{1}{2} a b \cos C}{\frac{1}{2} a b \sin C} \alpha=\alpha \cot C$