Use the Unit Circle to find [tex]\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)[/tex] in degrees. Remember that the domain of inverse sine is limited to quadrants I and IV (the right side of the unit circle).

A. [tex]225^{\circ}[/tex]
B. [tex]0^{\circ}[/tex]
C. [tex]60^{\circ}[/tex]
D. [tex]330^{\circ}[/tex]



Answer :

To find [tex]\(\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\)[/tex] using the Unit Circle, we must determine the angle whose sine value is [tex]\(\frac{\sqrt{3}}{2}\)[/tex].

### Step-by-Step Solution:

1. Understanding the Unit Circle:
- The unit circle is a circle with a radius of 1 centered at the origin of the coordinate plane.
- The sine of an angle in the unit circle corresponds to the y-coordinate of the point where the terminal side of the angle intersects the unit circle.

2. Inverse Sine Restriction:
- The domain of the inverse sine function, [tex]\(\sin^{-1}(x)\)[/tex], is [tex]\([-1, 1]\)[/tex], and its range is limited to [tex]\([-90^\circ, 90^\circ]\)[/tex], or in radians, [tex]\([- \frac{\pi}{2}, \frac{\pi}{2}]\)[/tex].
- This means the angles are limited to the first quadrant (0 to 90 degrees) and the fourth quadrant (-90 to 0 degrees).

3. Identifying the Relevant Angle:
- In the unit circle, an angle in the first quadrant where the sine value is [tex]\(\frac{\sqrt{3}}{2}\)[/tex] is at 60 degrees (or [tex]\(\frac{\pi}{3}\)[/tex] radians).
- We are interested in the principal value, which is the angle that falls within the range of the inverse sine function.

4. Matching the Angle with the Choices:
- Among the given choices:
- [tex]\(225^\circ\)[/tex] is in the third quadrant and does not have the sine value of [tex]\(\frac{\sqrt{3}}{2}\)[/tex].
- [tex]\(0^\circ\)[/tex] has a sine value of 0, not [tex]\(\frac{\sqrt{3}}{2}\)[/tex].
- [tex]\(330^\circ\)[/tex] is in the fourth quadrant and has a sine value of [tex]\(-\frac{\sqrt{3}}{2}\)[/tex], not [tex]\(\frac{\sqrt{3}}{2}\)[/tex].
- [tex]\(60^\circ\)[/tex] is in the first quadrant and has a sine value of [tex]\(\frac{\sqrt{3}}{2}\)[/tex].

Therefore, the angle [tex]\(\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\)[/tex] in degrees is:

[tex]\[\boxed{60^\circ}\][/tex]