`DeltaXYZ` is right angled at Y.If `mangleZ=30^(@)` ,then find the value of `sinx-(3)/(sqrt(2))`

Options

A. `((sqrt(6)-6))/(2sqrt(2))`
B. `((sqrt(2)-sqrt(3)))/(sqrt(6))`
C. `(-1)/(2)`
D. `((1-2sqrt(3)))/(2)`

Answer

To solve the problem, we need to find the value of \( \sin X - \frac{3}{\sqrt{2}} \) given that triangle \( \Delta XYZ \) is a right triangle with the right angle at \( Y \) and \( \angle Z = 30^\circ \).

### Step-by-Step Solution:

1. **Identify the Angles in Triangle XYZ**:
Since triangle \( XYZ \) is a right triangle at \( Y \), we know:
\[ \angle Y = 90^\circ \]
Given \( \angle Z = 30^\circ \), we can find \( \angle X \) using the property that the sum of angles in a triangle is \( 180^\circ \):
\[ \angle X + \angle Y + \angle Z = 180^\circ \]
Substituting the known values:
\[ \angle X + 90^\circ + 30^\circ = 180^\circ \]
Simplifying this gives:
\[ \angle X + 120^\circ = 180^\circ \]
Therefore:
\[ \angle X = 180^\circ - 120^\circ = 60^\circ \]

2. **Calculate \( \sin X \)**:
Now that we have \( \angle X = 60^\circ \), we can find \( \sin X \):
\[ \sin X = \sin 60^\circ \]
The value of \( \sin 60^\circ \) is:
\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \]

3. **Substitute into the Expression**:
We need to find:
\[ \sin X - \frac{3}{\sqrt{2}} \]
Substituting \( \sin X \):
\[ \frac{\sqrt{3}}{2} - \frac{3}{\sqrt{2}} \]

4. **Finding a Common Denominator**:
To subtract these fractions, we need a common denominator. The least common multiple of \( 2 \) and \( \sqrt{2} \) is \( 2\sqrt{2} \).
- For \( \frac{\sqrt{3}}{2} \):
\[ \frac{\sqrt{3}}{2} = \frac{\sqrt{3} \cdot \sqrt{2}}{2 \cdot \sqrt{2}} = \frac{\sqrt{6}}{2\sqrt{2}} \]
- For \( \frac{3}{\sqrt{2}} \):
\[ \frac{3}{\sqrt{2}} = \frac{3 \cdot 2}{\sqrt{2} \cdot 2} = \frac{6}{2\sqrt{2}} \]

5. **Combine the Fractions**:
Now we can combine the two fractions:
\[ \frac{\sqrt{6}}{2\sqrt{2}} - \frac{6}{2\sqrt{2}} = \frac{\sqrt{6} - 6}{2\sqrt{2}} \]

### Final Answer:
Thus, the value of \( \sin X - \frac{3}{\sqrt{2}} \) is:
\[ \frac{\sqrt{6} - 6}{2\sqrt{2}} \]