Let O be the center of the circle, and let AB be the chord subtending an angle of 48° at the center.
Since the angle subtended by the chord at the center is 48°, we know that the angle subtended by the same chord at the circumference is half of that, which is 24°.
We can draw a perpendicular from the center of the circle to the chord, which will bisect the chord and subtend a right angle at the point of intersection.
Let M be the midpoint of the chord AB, and let OM be the perpendicular from the center O to the midpoint M.
Since OM is perpendicular to the chord AB, we have formed two right-angled triangles OMA and OMB. Since OM bisects the chord, both triangles are congruent by the Side-Angle-Side (SAS) criterion.
Therefore, the length of the chord AB is 2 * AM.
In triangle OMA,
AM = OA * sin(24°) = 13 * sin(24°)
Using a calculator, we find that sin(24°) ≈ 0.4067.
Thus, AM = 13 * 0.4067 ≈ 5.2821 cm.
Therefore, the distance from the chord to the center of the circle is half the length of the chord, which is:
AB = 2 * AM = 2 * 5.2821 ≈ 10.5642 cm.
Therefore, the distance from the chord to the center of the circle is approximately 10.56 cm.
The angles subtended by a chord at the center of a circle of radius 13 cm is 48°. How far is the chord to the center of the circle
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