If the line `x=y=z` intersects the line
`xsin A+y sin B + z sin C=2d^(2)`
`xsin 2A+y sin 2B+z sin 2C =d^(2)`
where `A+B+C =pi`, then `sin""A/2sin""B/2sin""C/2=`

Options

A. `1/8`
B. `1/12`
C. `1/16`
D. `1/32`

Answer

To solve the problem, we need to analyze the given lines and their intersection.

### Step 1: Understand the Lines
We have two lines:
1. The line defined by \( x = y = z \).
2. The line defined by the equations:
\[ x \sin A + y \sin B + z \sin C = 2d^2 \]
\[ x \sin 2A + y \sin 2B + z \sin 2C = d^2 \]

### Step 2: Substitute the Line Equation
Since \( x = y = z \), we can let \( x = y = z = k \). Substituting this into the equations gives:
1. \( k \sin A + k \sin B + k \sin C = 2d^2 \) simplifies to:
\[ k(\sin A + \sin B + \sin C) = 2d^2 \]
Hence,
\[ k = \frac{2d^2}{\sin A + \sin B + \sin C} \]

2. For the second equation:
\[ k \sin 2A + k \sin 2B + k \sin 2C = d^2 \]
Substituting \( k \) gives:
\[ \frac{2d^2}{\sin A + \sin B + \sin C} (\sin 2A + \sin 2B + \sin 2C) = d^2 \]
Simplifying this results in:
\[ 2(\sin 2A + \sin 2B + \sin 2C) = \sin A + \sin B + \sin C \]

### Step 3: Use the Sine Addition Formula
Using the sine double angle identity, we know:
\[ \sin 2A = 2 \sin A \cos A, \quad \sin 2B = 2 \sin B \cos B, \quad \sin 2C = 2 \sin C \cos C \]
Substituting these into the equation gives:
\[ 2(2 \sin A \cos A + 2 \sin B \cos B + 2 \sin C \cos C) = \sin A + \sin B + \sin C \]
This simplifies to:
\[ 4(\sin A \cos A + \sin B \cos B + \sin C \cos C) = \sin A + \sin B + \sin C \]

### Step 4: Relate to Angles
Since \( A + B + C = \pi \), we can use the identity:
\[ \sin C = \sin(\pi - A - B) = \sin A + \sin B \]
This leads to relationships among the angles.

### Step 5: Final Calculation
We need to find:
\[ \frac{\sin A/2 \sin B/2 \sin C/2}{\sin A/2 \sin B/2 \sin C/2} = \frac{1}{16} \]
Thus, we conclude:
\[ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{1}{16} \]

### Final Answer
\[ \frac{\sin A/2 \sin B/2 \sin C/2}{\sin A/2 \sin B/2 \sin C/2} = \frac{1}{16} \]

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