In `DeltaABC,BC=1,sin.(A)/2=x_1,sin.(B)/2=x_2,cos.(A)/2=x_3andcos.(B)/2=x_4" with "(x_1/x_2)^2007-(x_3/x_4)^2006=0`
If `angleA=90^@`, then area of `DeltaABC` is

Options

A. 1/2 sq. units
B. 1/3 sq. units
C. 1 sq. units
D. 2sq. Units

Explanation

To find the area of triangle ABC given the conditions, we can follow these steps:

### Step 1: Understand the Triangle Configuration
Given that angle A = 90°, triangle ABC is a right-angled triangle with BC = 1. We need to determine the lengths of sides AB and AC.

### Step 2: Analyze the Given Conditions
We have:
- \( \frac{\sin(A/2)}{2} = x_1 \)
- \( \frac{\sin(B/2)}{2} = x_2 \)
- \( \frac{\cos(A/2)}{2} = x_3 \)
- \( \frac{\cos(B/2)}{2} = x_4 \)

Since angle A = 90°, we can find:
- \( A/2 = 45° \) so \( \sin(45°) = \frac{\sqrt{2}}{2} \) and \( \cos(45°) = \frac{\sqrt{2}}{2} \)

Thus,
- \( x_1 = \frac{\sqrt{2}/2}{2} = \frac{\sqrt{2}}{4} \)
- \( x_3 = \frac{\sqrt{2}/2}{2} = \frac{\sqrt{2}}{4} \)

### Step 3: Determine Angle B
Since triangle ABC is a right triangle and angle A = 90°, the sum of angles B and C must be 90°. Therefore, if we let angle B = x, then angle C = 90° - x.

### Step 4: Use the Given Condition
The condition given is:
\[ \left(\frac{x_1}{x_2}\right)^{2007} - \left(\frac{x_3}{x_4}\right)^{2006} = 0 \]
This implies:
\[ \frac{x_1}{x_2} = \frac{x_3}{x_4} \]
Since we have \( x_1 = x_3 \), we can conclude that \( x_2 = x_4 \).

### Step 5: Equate Angles
From the equality \( x_2 = x_4 \), we can conclude:
\[ \sin(B/2) = \cos(B/2) \]
This implies that \( B/2 = 45° \) or \( B = 90° \), which means angle B = 45° and angle C = 45°.

### Step 6: Determine the Lengths of Sides
In a 45°-45°-90° triangle, the lengths of the legs are equal. Let the lengths of AB and AC be x. By the Pythagorean theorem:
\[ AB^2 + AC^2 = BC^2 \]
Substituting the known values:
\[ x^2 + x^2 = 1^2 \]
\[ 2x^2 = 1 \implies x^2 = \frac{1}{2} \implies x = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

### Step 7: Calculate the Area
The area \( A \) of triangle ABC is given by:
\[ A = \frac{1}{2} \times AB \times AC = \frac{1}{2} \times x \times x = \frac{1}{2} \times \left(\frac{\sqrt{2}}{2}\right) \times \left(\frac{\sqrt{2}}{2}\right) \]
\[ A = \frac{1}{2} \times \frac{2}{4} = \frac{1}{4} \]

### Final Answer
The area of triangle ABC is \( \frac{1}{4} \).