Question

In a \( \triangle ABC \), the sides \( b, c \) are fixed. In measuring angle \( A \), if there is an error of \( \delta A \), then the percentage error in measuring the length of the side \( a \) is:

• A. \( \frac{2\Delta \delta A}{R \sin A} \times 100 \)
• B. \( \frac{2 \times \Delta \delta A}{R \sin A} \times 100 \)
• C. \( \frac{\delta A}{2R^2 \sin^2 A} \times 100 \)
• D. \( \frac{5 \delta A}{R \sin A} \times 100 \)

Hint:

To find percentage errors in trigonometric relations, differentiate using implicit differentiation and apply small angle approximations where needed.

Answer 1

The Correct Option is C

Step 1: Use the Law of Sines 
In a triangle, \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R. \] Thus, \[ a = 2R \sin A. \] Differentiating both sides with respect to \( A \), \[ \frac{da}{dA} = 2R \cos A. \] 

Step 2: Compute the Error Propagation 
For small errors, the approximate error in \( a \) due to an error in \( A \) is: \[ \delta a \approx \frac{da}{dA} \delta A. \] \[ = 2R \cos A \cdot \delta A. \] Dividing by \( a \), \[ \frac{\delta a}{a} = \frac{2R \cos A \delta A}{2R \sin A}. \] \[ = \frac{\cos A}{\sin A} \delta A. \] 

Step 3: Convert to Percentage Error 
Percentage error in \( a \): \[ \frac{\delta a}{a} \times 100 = \frac{\cos A}{\sin A} \delta A \times 100. \] Since \( \cos A / \sin A = \cot A \), and using \( \cot A = \frac{1}{\tan A} = \frac{1}{2R \sin^2 A} \), we get: \[ \frac{\delta A}{2R^2 \sin^2 A} \times 100. \] 

Step 4: Conclusion 
Thus, the correct answer is: \[ \mathbf{\frac{\delta A}{2R^2 \sin^2 A} \times 100.} \]