Find the angle formed by the line `sqrt(3)x+y-5=0` from the positive direction of `x`-axis.

Answer

To find the angle formed by the line given by the equation \( \sqrt{3}x + y - 5 = 0 \) from the positive direction of the x-axis, we can follow these steps:

### Step 1: Rearrange the equation to slope-intercept form
We start with the equation of the line:
\[ \sqrt{3}x + y - 5 = 0 \]
We can rearrange it to the form \( y = mx + c \):
\[ y = -\sqrt{3}x + 5 \]

### Step 2: Identify the slope
From the equation \( y = -\sqrt{3}x + 5 \), we can see that the slope \( m \) is:
\[ m = -\sqrt{3} \]

### Step 3: Relate the slope to the angle
The slope of a line is related to the angle \( \theta \) it makes with the positive x-axis by the formula:
\[ m = \tan(\theta) \]
Thus, we have:
\[ \tan(\theta) = -\sqrt{3} \]

### Step 4: Determine the angle
The value \( \tan(\theta) = -\sqrt{3} \) corresponds to angles where the tangent is negative. The reference angle where \( \tan(\theta) = \sqrt{3} \) is \( 60^\circ \). Since the tangent is negative, we look for angles in the second quadrant:
\[ \theta = 180^\circ - 60^\circ = 120^\circ \]

### Step 5: Conclusion
Therefore, the angle formed by the line from the positive direction of the x-axis is:
\[ \theta = 120^\circ \]