Consider the equation `sqrt(3)x+y-8=0`
I. Normal form of the given equation is
`cos 30^(@)x+sin 30^(@)y=4`
II. Values of p and w are 4 and `30^(@)` respectively.
Choose the correct option.

Options

A. Only I is true
B. Only II is true
C. Both are true
D. Both are false

Answer

To solve the given problem step by step, we will start from the equation provided and derive the normal form while identifying the values of \( p \) and \( w \).

### Step 1: Write the given equation
The given equation is:
\[ \sqrt{3}x + y - 8 = 0 \]

### Step 2: Rearrange the equation into the standard form
We can rearrange the equation to the form \( Ax + By + C = 0 \):
\[ \sqrt{3}x + y = 8 \]

### Step 3: Identify coefficients
From the equation \( \sqrt{3}x + y - 8 = 0 \), we can identify:
- \( A = \sqrt{3} \)
- \( B = 1 \)
- \( C = -8 \)

### Step 4: Calculate the slope of the line
The slope \( m \) of the line can be calculated as:
\[ m = -\frac{A}{B} = -\frac{\sqrt{3}}{1} = -\sqrt{3} \]

### Step 5: Find the angle \( \theta \)
The angle \( \theta \) that the line makes with the positive x-axis can be found using the tangent function:
\[ \tan \theta = -\sqrt{3} \]
This implies:
\[ \theta = 120^\circ \quad \text{(since tangent is negative in the second quadrant)} \]

### Step 6: Find the angle \( \alpha \)
Since the normal to the line makes an angle of \( 90^\circ \) with the line, we have:
\[ \alpha = 90^\circ - \theta = 90^\circ - 120^\circ = -30^\circ \]
However, we typically express angles in the range of \( [0^\circ, 180^\circ] \), so we can say:
\[ \alpha = 30^\circ \]

### Step 7: Calculate the distance \( p \)
The distance \( p \) from the origin (0, 0) to the line can be calculated using the formula:
\[ p = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Substituting \( (x_1, y_1) = (0, 0) \):
\[ p = \frac{|\sqrt{3}(0) + 1(0) - 8|}{\sqrt{(\sqrt{3})^2 + (1)^2}} = \frac{|-8|}{\sqrt{3 + 1}} = \frac{8}{2} = 4 \]

### Step 8: Write the normal form of the equation
The normal form of the equation is given by:
\[ \cos \alpha \cdot x + \sin \alpha \cdot y = p \]
Substituting \( \alpha = 30^\circ \) and \( p = 4 \):
\[ \cos 30^\circ \cdot x + \sin 30^\circ \cdot y = 4 \]
Using the values:
\[ \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sin 30^\circ = \frac{1}{2} \]
Thus, the equation becomes:
\[ \frac{\sqrt{3}}{2}x + \frac{1}{2}y = 4 \]

### Step 9: Multiply through by 2 to eliminate fractions
\[ \sqrt{3}x + y = 8 \]

### Conclusion
The normal form of the given equation is:
\[ \cos 30^\circ \cdot x + \sin 30^\circ \cdot y = 4 \]
with \( p = 4 \) and \( w = 30^\circ \).

### Final Answer
Both statements I and II are correct.