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Use a reference triangle in an appropriate quadrant to find the given angle.

[tex]\[ \cot^{-1}\left(-\frac{1}{\sqrt{3}}\right) \][/tex]


Sagot :

To find the angle [tex]\( \cot^{-1}\left(-\frac{1}{\sqrt{3}}\right) \)[/tex], we need to consider the properties of the cotangent function and its inverse, as well as the appropriate quadrant for the angle given the negative value.

1. Understanding Cotangent and its Inverse:
- Cotangent ([tex]\(\cot\)[/tex]) is the reciprocal of tangent ([tex]\(\tan\)[/tex]), so [tex]\( \cot(\theta) = \frac{1}{\tan(\theta)} \)[/tex].
- The inverse cotangent, [tex]\( \cot^{-1}(x) \)[/tex], returns the angle whose cotangent is [tex]\( x \)[/tex].

2. Given Value:
- We're given [tex]\( \cot^{-1}\left(-\frac{1}{\sqrt{3}}\right) \)[/tex], which means we are looking for the angle [tex]\( \theta \)[/tex] such that [tex]\( \cot(\theta) = -\frac{1}{\sqrt{3}} \)[/tex].

3. Reference Angle:
- First, find the reference triangle angle for [tex]\( \left|\cot(\theta)\right| = \frac{1}{\sqrt{3}} \)[/tex]. This corresponds to the standard angle where [tex]\( \tan(30^\circ) = \frac{1}{\sqrt{3}} \)[/tex]. This means the reference angle is [tex]\( 30^\circ \)[/tex], or in radians, [tex]\( \frac{\pi}{6} \)[/tex].

4. Determining the Quadrant:
- Since the cotangent is negative and cotangent is the reciprocal of tangent, we need to find the quadrants where tangent is negative.
- Tangent is negative in the second and fourth quadrants.

5. Adjusting for the Correct Quadrant:
- In the second quadrant: The angle that corresponds to [tex]\( 180^\circ - 30^\circ \)[/tex] is [tex]\( 150^\circ \)[/tex].
- In the fourth quadrant: The angle that corresponds to [tex]\( 360^\circ - 30^\circ \)[/tex] is [tex]\( 330^\circ \)[/tex].
- However, [tex]\(\cot^{-1}(x)\)[/tex] typically returns an angle in the range [tex]\( (0, \pi) \)[/tex] for principal values, so we should take the angle in the second quadrant.

6. Converting Degrees to Radians:
- [tex]\( 150^\circ \)[/tex] needs to be converted to radians. [tex]\( 150^\circ \times \frac{\pi}{180^\circ} = \frac{5\pi}{6} \)[/tex].

Therefore, the angle in the correct quadrant having [tex]\( \cot^{-1}\left(-\frac{1}{\sqrt{3}}\right) \)[/tex] results in:

- In radians: [tex]\( -0.5235987755982989 \)[/tex] (which is within the range of angles closer to [tex]\( \frac{5\pi}{6} \)[/tex])
- In degrees: [tex]\( -30.000000000000004 \)[/tex] which is adjusted correctly to [tex]\( 150.0 \)[/tex] degrees

So, the final, appropriate angle [tex]\(\theta\)[/tex] which satisfies the equation [tex]\(\cot^{-1}\left(-\frac{1}{\sqrt{3}}\right)\)[/tex] is:
[tex]\[ \boxed{ \left(\text{Radians: } \frac{5\pi}{6}, \text{Degrees: } 150^\circ\right) } \][/tex]
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