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Chapter 6: Problem 22

Find the inclination \(\theta\) (in radians and degrees) of the line passing through the points. \((3, \sqrt{3}),(6,-2 \sqrt{3})\)

Short Answer

Expert verified
The inclination angle of the line passing through the points \((3, \sqrt{3})\) and \((6,-2\sqrt{3})\) is \(-\frac{\pi}{3}\) radians or \(-60\) degrees.

Step by step solution

01

Calculate the Slope

Find the slope of the line formed by the points using the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\). For the points \((3, \sqrt{3})\), and \((6,-2 \sqrt{3})\), the slope \(m\) will be \(\frac{-2\sqrt{3} - \sqrt{3}}{6 - 3} = -\sqrt{3}\).
02

Find the Inclination using arctan

The angle of inclination \(\theta\) of the line can be found using the arctan function applied on the slope. So, \(\theta = \arctan(m) = \arctan(-\sqrt{3})\). Using the standard arctan values, \(\arctan(-\sqrt{3}) = -\frac{pi}{3}\).
03

Convert the Inclination to degrees

To get the angle in degrees, we convert \(\theta\) from radians to degrees by multiplying it by \(\frac{180}{\pi}\). Hence, the angle in degrees is \(-\frac{pi}{3} \times \frac{180}{\pi} = -60 \) degrees.

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