In a ` Delta` ABC if sides a and b remain constant such that a is the error in C, thenrelatinv error in its area, is

Options

A. `a cot C `
B. a sin C
C. a tan C
D. a cos C

Explanation

To solve the problem, we need to find the relative error in the area of triangle ABC when the sides a and b remain constant, and the error in angle C is denoted as \( \Delta C \).

### Step-by-Step Solution:

1. **Understand the Area Formula**:
The area \( S \) of triangle ABC can be expressed using the formula:
\[ S = \frac{1}{2}ab \sin C \]
where \( a \) and \( b \) are the lengths of the sides, and \( C \) is the angle between them.

**Hint**: Recall the formula for the area of a triangle using two sides and the sine of the included angle.

2. **Differentiate the Area with Respect to Angle C**:
To find how the area changes with respect to angle \( C \), we differentiate \( S \) with respect to \( C \):
\[ \frac{dS}{dC} = \frac{1}{2}ab \cos C \]

**Hint**: Use the chain rule for differentiation, noting that the derivative of \( \sin C \) is \( \cos C \).

3. **Relate the Change in Area to the Change in Angle**:
The change in area \( \Delta S \) can be expressed in terms of the change in angle \( \Delta C \):
\[ \Delta S = \frac{dS}{dC} \Delta C \]
Substituting the expression for \( \frac{dS}{dC} \):
\[ \Delta S = \left(\frac{1}{2}ab \cos C\right) \Delta C \]

**Hint**: Recognize that \( \Delta C \) is the error in angle \( C \), which is given as \( A \) in the problem.

4. **Substitute the Error in Angle**:
Since \( \Delta C = A \), we can substitute this into our equation:
\[ \Delta S = \frac{1}{2}ab \cos C \cdot A \]

**Hint**: Make sure to replace \( \Delta C \) with the given error \( A \).

5. **Calculate the Relative Error in Area**:
The relative error in the area is given by:
\[ \text{Relative Error} = \frac{\Delta S}{S} \]
Substituting \( S \) and \( \Delta S \):
\[ \text{Relative Error} = \frac{\frac{1}{2}ab \cos C \cdot A}{\frac{1}{2}ab \sin C} \]
The \( \frac{1}{2}ab \) cancels out:
\[ \text{Relative Error} = \frac{\cos C \cdot A}{\sin C} = A \cot C \]

**Hint**: Simplify the fraction by canceling out common terms.

### Final Result:
The relative error in the area of triangle ABC is:
\[ \text{Relative Error} = A \cot C \]