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Chapter 6: Problem 19

Find the inclination \(\theta\) (in radians and degrees) of the line passing through the points. \((\sqrt{3}, 2),(0,1)\)

Short Answer

Expert verified
The inclination of the line passing through the points \((\sqrt{3}, 2)\) and \((0,1)\) is \(-\frac{\pi}{6}\) radians or \(-30^{\circ}\).

Step by step solution

01

Determining the Slope of the Line

Identify the coordinates of the two points as they are given. The first point has coordinates \((\sqrt{3}, 2)\) and will be labeled as (x1, y1). The second point has coordinates \((0, 1)\) and will be labeled as (x2, y2). We will determine the slope m of the line passing through these points. The formula for the slope is \(m = \frac{y2-y1}{x2-x1}\).Substituting these values into the slope formula gives us the slope \(m = \frac{1-2}{0-\sqrt{3}} = -\frac{1}{\sqrt{3}}\).
02

Computing the Inclination in Radians

The inclination of a line is the angle that it makes with the positive x-axis. We can find this inclination by taking the arctangent of the slope (m):\(\theta = arctan(m) = arctan\left(-\frac{1}{\sqrt{3}}\right)\).By using the properties of the arctangent and the given slope, we find the inclination to be \(\theta = -\frac{\pi}{6}\) radians.
03

Converting Radians to Degrees

Finally, we will convert the radians into degrees using the relation \(1 rad = \frac{180}{\pi}^{\circ}\). This gives us \(\theta = -\frac{\pi}{6} * \frac{180}{\pi} = -30^{\circ}\).

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