A train is moving at a constant rate at an angle `theta` East of North. Observations of the train are made from a fixed point. It is due north at some instant. Ten minutes earlier its bearing was `alpha_(1)` West of North whereas ten minutes afterwards its bearing `alpha_(2)` East of North, then `tantheta=`____.

Answer

To solve the problem step by step, we will analyze the situation involving the train's movement and the angles of observation.

### Step 1: Understanding the Problem
We have a train moving at a constant angle \( \theta \) East of North. At a certain moment, the train is directly North of a fixed observation point \( A \). Ten minutes earlier, its bearing was \( \alpha_1 \) West of North, and ten minutes later, its bearing was \( \alpha_2 \) East of North.

### Step 2: Setting Up the Diagram
1. Draw a point \( A \) representing the observation point.
2. Mark the position of the train at the current moment as point \( O \) (due North of \( A \)).
3. Mark the position of the train ten minutes earlier as point \( P \) (bearing \( \alpha_1 \) West of North).
4. Mark the position of the train ten minutes later as point \( Q \) (bearing \( \alpha_2 \) East of North).

### Step 3: Identifying Angles
- The angle \( \angle PAO \) is \( \alpha_1 \) (West of North).
- The angle \( \angle QAO \) is \( \alpha_2 \) (East of North).
- The angle \( \theta \) represents the angle of the train's movement relative to North.

### Step 4: Establishing Relationships
Since the train moves at a constant speed, the distances \( OP \) and \( OQ \) covered in the same time interval (10 minutes) are equal. Let this distance be \( d \).

### Step 5: Applying Trigonometric Relationships
From triangle \( PAO \):
- The horizontal distance to point \( P \) can be expressed as \( d \sin(\alpha_1) \).
- The vertical distance to point \( P \) can be expressed as \( d \cos(\alpha_1) \).

From triangle \( QAO \):
- The horizontal distance to point \( Q \) can be expressed as \( d \sin(\alpha_2) \).
- The vertical distance to point \( Q \) can be expressed as \( d \cos(\alpha_2) \).

### Step 6: Using Trigonometric Identities
Using the cotangent function:
- \( \cot(\theta) = \frac{\text{adjacent}}{\text{opposite}} \)

From the triangles:
- For triangle \( PAO \):
\[ d \cot(\alpha_1) = d \cos(\alpha_1) / d \sin(\alpha_1) \]
- For triangle \( QAO \):
\[ d \cot(\alpha_2) = d \cos(\alpha_2) / d \sin(\alpha_2) \]

### Step 7: Setting Up the Equation
Since the distances \( OP \) and \( OQ \) are equal:
\[ d \cot(\alpha_2) - d \cot(\alpha_1) = d \cot(\theta) \]
This simplifies to:
\[ \cot(\theta) = \frac{1}{2} (\cot(\alpha_2) - \cot(\alpha_1)) \]

### Step 8: Finding \( \tan(\theta) \)
Using the relationship between cotangent and tangent:
\[ \tan(\theta) = \frac{1}{\cot(\theta)} = \frac{2}{\cot(\alpha_2) - \cot(\alpha_1)} \]

### Step 9: Final Expression
Using the sine and cosine definitions:
\[ \tan(\theta) = \frac{2 \sin(\alpha_1) \sin(\alpha_2)}{\sin(\alpha_1 - \alpha_2)} \]

### Conclusion
Thus, we have derived the expression for \( \tan(\theta) \) as:
\[ \tan(\theta) = \frac{2 \sin(\alpha_1) \sin(\alpha_2)}{\sin(\alpha_1 - \alpha_2)} \]