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We are given the formula for the area \( A \) of a triangle with sides \( b \), \( c \), and included angle \( \theta \):

\[
A = \frac{1}{2} bc \sin(\theta)
\]

The task is to calculate the differential \( dA \), and then use the relative error formula \( \frac{dA}{A} \) to estimate the maximum tolerable error in measuring the angle \( \theta \).

### Step 1: Differential of \( A \)

The differential of \( A \) is found by taking the total derivative of the formula \( A = \frac{1}{2}bc\sin(\theta) \):

\[
dA = \frac{\partial A}{\partial b} db + \frac{\partial A}{\partial c} dc + \frac{\partial A}{\partial \theta} d\theta
\]

Now calculate each partial derivative:

- \( \frac{\partial A}{\partial b} = \frac{1}{2} c \sin(\theta) \)
- \( \frac{\partial A}{\partial c} = \frac{1}{2} b \sin(\theta) \)
- \( \frac{\partial A}{\partial \theta} = \frac{1}{2} bc \cos(\theta) \)

Thus, the differential \( dA \) is:

\[
dA = \frac{1}{2} c \sin(\theta) db + \frac{1}{2} b \sin(\theta) dc + \frac{1}{2} bc \cos(\theta) d\theta
\]

### Step 2: Relative Error

The relative error \( \frac{dA}{A} \) can now be written as:

\[
\frac{dA}{A} = \frac{\frac{1}{2} c \sin(\theta) db + \frac{1}{2} b \sin(\theta) dc + \frac{1}{2} bc \cos(\theta) d\theta}{\frac{1}{2} bc \sin(\theta)}
\]

This simplifies to:

\[
\frac{dA}{A} = \frac{db}{b} + \frac{dc}{c} + \frac{\cos(\theta)}{\sin(\theta)} d\theta
\]

\[
\frac{dA}{A} = \frac{db}{b} + \frac{dc}{c} + \cot(\theta) d\theta
\]

### Step 3: Maximum Error

We are given that:

- Relative error in area \( \frac{dA}{A} \leq 0.02 \) (2%),
- Relative error in lengths \( \frac{db}{b}, \frac{dc}{c} \leq 0.005 \) (0.5%).

Plugging these values into the relative error equation:

\[
0.02 \geq 0.005 + 0.005 + \cot(\theta) d\theta
\]

This simplifies to:

\[
0.02 \geq 0.01 + \cot(\theta) d\theta
\]

Thus:

\[
0.01 \geq \cot(\theta) d\theta
\]

Solving for \( d\theta \):

\[
d\theta \leq \frac{0.01}{\cot(\theta)} = 0.01 \tan(\theta)
\]

### Step 4: Maximum Tolerable Error in Degrees

The maximum tolerable error \( d\theta \) will depend on \( \theta \), the angle. Let's calculate \( d\theta \) for \( \theta = 10^\circ \) and \( \theta = 24^\circ \).

- For \( \theta = 10^\circ \), \( \tan(10^\circ) \approx 0.1763 \), so:

\[
d\theta \leq 0.01 \times 0.1763 \approx 0.00176 \, \text{radians} = 0.101^\circ
\]

- For \( \theta = 24^\circ \), \( \tan(24^\circ) \approx 0.4452 \), so:

\[
d\theta \leq 0.01 \times 0.4452 \approx 0.00445 \, \text{radians} = 0.255^\circ
\]

Thus, the maximum tolerable error in measuring angles is between approximately \( 0.101^\circ \) and \( 0.255^\circ \), depending on the angle.

---

Would you like a more detailed breakdown of any of the steps?

Here are some related questions you might find helpful:

1. What is the significance of using differentials in error analysis?
2. How does the choice of angle range (10° to 24°) affect the error tolerance?
3. What happens if the tolerable error in lengths is increased or decreased?
4. How is the error tolerance in radians converted into degrees?
5. What are some practical applications of error analysis in geometry?

**Tip**: When working with trigonometric functions in error analysis, always make sure to express small angle errors in radians before converting them to degrees.

We wish to compute the area of triangles with angles between 10° and 24° with a relative error of no more than 2%. Lengths can be measured with a relative error of no more than 0.5%. Use the expression for dA/A to estimate the maximum absolute error we can tolerate in the measurement of angles.

Learn how to calculate the maximum tolerable error in measuring angles for triangles with side lengths and angles using trigonometric and differential calculus methods. This problem covers key concepts such as error propagation, differentials, and trigonometric functions. Suitable for advanced high school or undergraduate students.

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