If the bisector of angle `A` of the triangle `ABC` makes an angle `theta` with `BC`, then `sintheta=`

Options

A. `|sin((B-C)/(2))|`
B. `|sin((B)/(2)-C)|`
C. `cos((B-C)/(2))`
D. `cos((B)/(2)-C)`

Explanation

To find the value of \( \sin \theta \) where \( \theta \) is the angle made by the bisector of angle \( A \) of triangle \( ABC \) with side \( BC \), we can follow these steps:

### Step 1: Understand the triangle and the angle bisector
In triangle \( ABC \), let \( A \) be the angle at vertex \( A \), \( B \) at vertex \( B \), and \( C \) at vertex \( C \). The angle bisector of angle \( A \) divides it into two equal angles, each measuring \( \frac{A}{2} \).

### Step 2: Set up the angle relationship
Since the sum of angles in a triangle is \( 180^\circ \), we can express angle \( A \) in terms of angles \( B \) and \( C \):
\[ A + B + C = 180^\circ \implies A = 180^\circ - B - C \]

### Step 3: Substitute into the angle bisector equation
Now, substituting for \( A \):
\[ \frac{A}{2} + \theta + C = 180^\circ \]
Substituting \( A \):
\[ \frac{180^\circ - B - C}{2} + \theta + C = 180^\circ \]

### Step 4: Simplify the equation
Multiply through by 2 to eliminate the fraction:
\[ 180^\circ - B - C + 2\theta + 2C = 360^\circ \]
This simplifies to:
\[ 2\theta - B + C = 180^\circ \]
Rearranging gives:
\[ 2\theta = 180^\circ + B - C \implies \theta = 90^\circ + \frac{B - C}{2} \]

### Step 5: Find \( \sin \theta \)
Now we can find \( \sin \theta \):
\[ \sin \theta = \sin\left(90^\circ + \frac{B - C}{2}\right) \]
Using the identity \( \sin(90^\circ + x) = \cos x \):
\[ \sin \theta = \cos\left(\frac{B - C}{2}\right) \]

### Final Result
Thus, the required expression for \( \sin \theta \) is:
\[ \sin \theta = \cos\left(\frac{B - C}{2}\right) \]