Circle Chord and Angle Relationship

The question asks about the relationship between the radius of a circle and the length of a chord AB that subtends an angle of $30^\circ$ at the circumference.

Let O be the center of the circle and r be its radius. The chord is AB. The angle subtended by the chord AB at a point on the circumference is given as $30^\circ$. Let this point on the circumference be C, so $\angle ACB = 30^\circ$.

Chord Subtends Angle at Center

There is a fundamental theorem in circle geometry which states that the angle subtended by an arc at the center is double the angle subtended by the same arc at any point on the remaining part of the circle's circumference.

In our case, the chord AB corresponds to an arc AB. The angle subtended by arc AB at the circumference is $\angle ACB = 30^\circ$. The angle subtended by the same arc AB at the center O is $\angle AOB$.

According to the theorem:

$\angle AOB = 2 \times \angle ACB$

$\angle AOB = 2 \times 30^\circ$

$\angle AOB = 60^\circ$

So, the chord AB subtends an angle of $60^\circ$ at the center of the circle.

Triangle Formed by Radius and Chord

Consider the triangle $\triangle AOB$. The vertices are the center O and the endpoints of the chord A and B. The sides OA and OB are radii of the circle, so OA = OB = r.

Since OA = OB, $\triangle AOB$ is an isosceles triangle. In an isosceles triangle, the angles opposite the equal sides are equal. Thus, $\angle OAB = \angle OBA$.

The sum of angles in any triangle is $180^\circ$. In $\triangle AOB$:

$\angle OAB + \angle OBA + \angle AOB = 180^\circ$

Since $\angle OAB = \angle OBA$ and $\angle AOB = 60^\circ$, we can write:

$\angle OAB + \angle OAB + 60^\circ = 180^\circ$

$2 \angle OAB = 180^\circ - 60^\circ$

$2 \angle OAB = 120^\circ$

$\angle OAB = \frac{120^\circ}{2}$

$\angle OAB = 60^\circ$

Since $\angle OAB = \angle OBA$, we also have $\angle OBA = 60^\circ$.

Radius and Chord Length Comparison

In $\triangle AOB$, we found that:

• $\angle OAB = 60^\circ$
• $\angle OBA = 60^\circ$
• $\angle AOB = 60^\circ$

Since all three angles of $\triangle AOB$ are $60^\circ$, it is an equilateral triangle.

In an equilateral triangle, all sides are equal in length. Therefore:

OA = OB = AB

Since OA and OB are radii (r), and AB is the chord length, we have:

r = AB

This means the radius of the circle is the same as the length of the chord AB.

Conclusion on Radius vs. Chord Length

Based on our analysis, when a chord subtends an angle of $30^\circ$ at the circumference, the triangle formed by the chord and the radii to its endpoints is equilateral. This directly implies that the length of the chord is equal to the radius of the circle.

Comparing this finding with the given options:

• independent of length AB: Incorrect
• the same as length AB: Correct
• larger than length AB: Incorrect
• smaller than length AB: Incorrect