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
Answer & explanation
Correct answer: option 1
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,
$\Delta S =\frac{d S}{d C} \Delta C$ $[∵ \Delta C=\alpha]$
$\Rightarrow \Delta S =\frac{1}{2} a b ~\cos C ~\alpha$
$\Rightarrow \frac{\Delta S}{S}=\frac{\frac{1}{2} a b \cos C}{\frac{1}{2} a b \sin C} \alpha=\alpha \cot C$