Understanding the Circle Geometry Problem

The problem asks for the length of a chord in a unit circle when the chord subtends an angle of 120° at the center. A unit circle is a circle with a radius of 1 unit. The angle subtended at the center by the chord is given as 120°.

Let the circle have its center at O and the chord be AB. The radius of the circle is OA = OB = 1 unit. The angle \(\angle AOB = 120^\circ\).

Calculating the Chord Length

We can find the length of the chord AB using trigonometry. Consider the triangle formed by the center and the endpoints of the chord, triangle AOB.

Triangle AOB is an isosceles triangle because OA and OB are radii of the same circle (OA = OB = 1). We can use the Law of Cosines in triangle AOB to find the length of side AB (the chord).

Using the Law of Cosines

The Law of Cosines states that for any triangle with sides a, b, c and angle C opposite side c:

\(c^2 = a^2 + b^2 - 2ab \cos(C)\)

In our triangle AOB, let AB = c, OA = a = 1, OB = b = 1, and \(\angle AOB = C = 120^\circ\). Substituting these values:

\(AB^2 = OA^2 + OB^2 - 2(OA)(OB) \cos(\angle AOB)\)

\(AB^2 = 1^2 + 1^2 - 2(1)(1) \cos(120^\circ)\)

We know that \(\cos(120^\circ) = \cos(180^\circ - 60^\circ) = -\cos(60^\circ) = -\frac{1}{2}\).

Substitute the value of \(\cos(120^\circ)\):

\(AB^2 = 1 + 1 - 2(1)(1)\left(-\frac{1}{2}\right)\)

\(AB^2 = 2 - 2\left(-\frac{1}{2}\right)\)

\(AB^2 = 2 - (-1)\)

\(AB^2 = 2 + 1\)

\(AB^2 = 3\)

To find AB, we take the square root of both sides:

\(AB = \sqrt{3}\)

So, the length of the chord is \(\sqrt{3}\) units.

Alternative Method: Using a Perpendicular Bisector

Draw a perpendicular line from the center O to the chord AB. Let this line intersect the chord at point M. In an isosceles triangle AOB, the altitude OM from O to AB bisects both the angle \(\angle AOB\) and the chord AB.

So, \(\angle AOM = \angle BOM = \frac{120^\circ}{2} = 60^\circ\). Also, AM = MB = \(\frac{AB}{2}\).

Consider the right-angled triangle AOM. We have:

• Hypotenuse OA = 1 (radius of the unit circle).
• Angle \(\angle AOM = 60^\circ\).
• Side AM is opposite to angle \(\angle AOM\).

Using the sine function in triangle AOM:

\(\sin(\angle AOM) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AM}{OA}\)

\(\sin(60^\circ) = \frac{AM}{1}\)

We know that \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\).

\(\frac{\sqrt{3}}{2} = \frac{AM}{1}\)

So, \(AM = \frac{\sqrt{3}}{2}\).

The length of the chord AB is \(2 \times AM\):

\(AB = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}\)

Both methods confirm that the length of the chord is \(\sqrt{3}\) units.

Comparing with Options

Let's compare our calculated chord length with the given options:

• Option 1: \(\sqrt{2} - 1\) units
• Option 2: \(\sqrt{3} - 1\) units
• Option 3: \(\sqrt{2}\) units
• Option 4: \(\sqrt{3}\) units

Our calculated length \(\sqrt{3}\) matches Option 4.

Summary of Calculation Steps

1. Identify the given information: Unit circle (radius = 1), Chord subtends angle 120° at the center.
2. Recognize the triangle formed by the center and chord endpoints is isosceles.
3. Use either the Law of Cosines or the perpendicular bisector method.
4. Apply the relevant trigonometric or geometric principles (\(\cos(120^\circ)\) or \(\sin(60^\circ)\)).
5. Calculate the length of the chord.
6. Compare the result with the given options.

Revision Table: Key Formulas in Circle Geometry

Additional Information: Unit Circle Properties and Special Angles

A unit circle is fundamental in trigonometry. It's centered at the origin (0,0) in the Cartesian plane and has a radius of 1. Points on the unit circle have coordinates \((x, y) = (\cos(\theta), \sin(\theta))\), where \(\theta\) is the angle measured counterclockwise from the positive x-axis.

Understanding the sine and cosine values for special angles like 30°, 45°, 60°, 90°, and their related angles in other quadrants (like 120°, 150°, etc.) is crucial for solving many geometry and trigonometry problems.

• \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\)
• \(\cos(60^\circ) = \frac{1}{2}\)
• \(\cos(120^\circ) = -\frac{1}{2}\) (since 120° is in the second quadrant where cosine is negative)

These values were directly used in calculating the chord length in the unit circle problem.