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(D) Let $	heta = A = 67 \frac{1}{2}^{\circ} = \frac{135^{\circ}}{2} = \frac{3\pi}{8} 	ext{ radians}$.Given the error in measurement $d	heta = 9 	e

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

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(D) Let $	heta = A = 67 \frac{1}{2}^{\circ} = \frac{135^{\circ}}{2} = \frac{3\pi}{8} 	ext{ radians}$.Given the error in measurement $d	heta = 9 	ext{ min} = \frac{9}{60}^{\circ} = \frac{3}{20}^{\circ} = \frac{3}{20} 	imes \frac{\pi}{180} 	ext{ radians} = \frac{\pi}{1200} 	ext{ radians}$.The area of the triangle is $S = \frac{1}{2}bc \sin 	heta$.Differentiating with respect to $	heta$,we get $\frac{dS}{d	heta} = \frac{1}{2}bc \cos 	heta$.Thus,the absolute error in area is $dS = \frac{1}{2}bc \cos 	heta d	heta$.The relative error is $\frac{dS}{S} = \frac{\frac{1}{2}bc \cos 	heta d	heta}{\frac{1}{2}bc \sin 	heta} = \cot 	heta d	heta$.Substituting the values,$\frac{dS}{S} = \cot \left( \frac{3\pi}{8} ight) 	imes \frac{\pi}{1200}$.Since $\cot \left( \frac{3\pi}{8} ight) = 	an \left( \frac{\pi}{8} ight) = \sqrt{2}-1$,we have $\frac{dS}{S} = (\sqrt{2}-1) \frac{\pi}{1200}$.To express this as a percentage,we multiply by $100$: $	ext{Percentage error} = (\sqrt{2}-1) \frac{\pi}{1200} 	imes 100 = \frac{\pi}{12}(\sqrt{2}-1)$.

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

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