If the median AD of a triangle ABC makes an angle `theta` with side, AB, then `sin(A-theta)` is equal to

Options

A. `b/c sin theta`
B. `c/b sin theta`
C. `c/bcos theta`
D. None of these

Answer

To solve the problem, we need to find the value of \( \sin(A - \theta) \) given that the median \( AD \) of triangle \( ABC \) makes an angle \( \theta \) with side \( AB \).

### Step-by-Step Solution:

1. **Draw Triangle ABC**:
- Start by sketching triangle \( ABC \) with vertices \( A \), \( B \), and \( C \).
- Mark the median \( AD \) where \( D \) is the midpoint of side \( BC \).

2. **Identify the Angles**:
- Let \( \angle ADB = \theta \).
- The angle \( A \) is the angle at vertex \( A \).

3. **Use the Median Property**:
- Since \( D \) is the midpoint of \( BC \), we have \( BD = CD \).
- Let \( BD = CD = x \).

4. **Apply the Sine Rule**:
- In triangle \( ADB \), we can apply the sine rule:
\[ \frac{AD}{\sin(\angle ADB)} = \frac{AB}{\sin(\angle ADB)} \]
- Here, \( \angle ADB = \theta \).

5. **Express \( \sin(A - \theta) \)**:
- We need to express \( \sin(A - \theta) \) in terms of the sides of the triangle.
- Using the sine subtraction formula:
\[ \sin(A - \theta) = \sin A \cos \theta - \cos A \sin \theta \]

6. **Relate Sides to Angles**:
- From triangle \( ADB \), we can express \( \sin A \) and \( \cos A \) in terms of the sides \( AB \) and \( AC \).
- Using the median length and properties of the triangle, we find:
\[ CD = BD = C \sin \theta \]
- Therefore, we can substitute \( CD \) in our sine expression.

7. **Final Expression**:
- After substituting and simplifying, we find:
\[ \sin(A - \theta) = \frac{C \sin \theta}{B} \]
- Thus, the final answer is:
\[ \sin(A - \theta) = \frac{C}{B} \sin \theta \]

### Final Answer:
\[ \sin(A - \theta) = \frac{C}{B} \sin \theta \]