The angle $A$ of $\triangle A B C$ is found by measurement to be $67 \frac{1^{\circ}}{2}$ and the area of $\triangle A B C$ is calculated from the measurements of $b, c, A$. In measuring $A$, an error of $9 \mathrm{~min}$ is made then the percentage error in the area of the triangle is

• A

  $\frac{\pi}{6}(2-\sqrt{3})$
• B

  $\frac{\pi}{6}(2+\sqrt{3})$
• C

  $\frac{\pi}{12}(\sqrt{2}+1)$
• Correct answer
  D

  $\frac{\pi}{12}(\sqrt{2}-1)$

Solution

Given, $d \theta=9 \mathrm{~min}$ In radian $d \theta=\frac{9}{60} \times \frac{\pi}{180}=\frac{\pi}{1200}$
$\begin{aligned} A =\frac{1}{2} b c \sin \theta \Rightarrow \frac{d A}{d \theta}=\frac{1}{2} b c \cos \theta \Rightarrow d A=\frac{1}{2} b c \cos \theta d \theta \\ \frac{d A}{A} \times 100=\frac{\frac{1}{2} b c \cos \theta d \theta}{\frac{1}{2} b c \sin \theta} \times 100=\cot \theta d \theta \times 100 \\ =\cot \frac{3 \pi}{8} \times \frac{\pi}{1200} \times 100 {\left[\because \theta=67 \frac{1}{2}^{\circ}=\frac{3 \pi}{8}\right] } \\ =(\sqrt{2}-1) \frac{\pi}{12} {\left[\because \cos \frac{3 \pi}{8}=\sqrt{2}-1\right] }\end{aligned}$

Hence, the correct answer is option: 'D'.