The diameter AB and chord CD of a circle with a radius of 4 cm intersect at right angles.

The diameter AB and chord CD of a circle with a radius of 4 cm intersect at right angles. calculate the length of the chord CD if the CAD angle = 60 degrees.

By condition, the diameter AB and the chord CD intersect at right angles, and from the properties of the chord it is known that the diameter of a circle perpendicular to the chord divides this chord in half. AB intersects CD at point K, then the segments CK and KD are equal.
Consider triangle CAD: СK = KD, AK – height and median. If in a triangle the median and the height coincide, then this triangle is isosceles. Therefore, AC and AD are the sides of the isosceles triangle CAD, CD is the base, and the angles ACD and ADC are the angles at the base. The angles at the base of an isosceles triangle are equal, then:
angle ACD = angle ADC = x.
By the theorem on the sum of the angles of a triangle:
CAD angle + ACD angle + ADC angle = 180 degrees;
60 + x + x = 180;
2x = 180 – 60;
2x = 120;
x = 120/2;
x = 60.
Thus, angle ACD = angle ADC = x = 60 degrees.
Since all angles in the CAD triangle are 60 degrees, then this triangle is equilateral (regular), inscribed in a circle with a radius of 4 cm, then CA = AD = CD.
The length of the radius of a circle circumscribed about a regular triangle is found by the formula:
R = √3a / 3,
where a is the side of the triangle.
Substitute the known values:
√3a / 3 = 4;
√3а = 12 (proportional);
a = 12 / √3 (proportional);
a = 12√3 / 3;
a = 4√3.
CA = AD = CD = a = 4√3 cm.
Answer: CD = 4√3 cm.