The given figure shows a circle with center O, AB and CD are two chords of the circle which are at a distance of 5 cm from the center O and ∠ABO = 40° CD = 24 cm.

The measure of the angle subtended by the chord CD at the center of the circle, and the radius of circle are respectively:

It is given that; the chords AB and CD are equidistant from the center O.

∴ AB = CD [Since chords of a circle which are equidistant from the centre are equal in length]

∴AB = 24 cm [Given that, CD = 24 cm.]

Draw OE ⊥ AB,

Then, OE = 5 cm.

We know that, the perpendicular from the centre of a circle to a chord bisects the chord.

∴AE = EB = $\frac{1}{2}$AB =$\frac{1}{2}$×24 cm = 12 cm

In right triangle ΔOAE, by Pythagorean relationship, we have:

OA² = AE² + OE²

= 12² + 5² (Perpendicular from center bisects the chords)

= 144 + 25

= 169

∴OA = $\sqrt{169}$ cm = 13 cm

Now, in ΔAOB

∠A = ∠B [Since, ΔAOB is an isosceles triangle as OA = OB (both are radii)]

∴∠A = 40°

∴∠AOB = 180° − (∠A + ∠B)

= 180° − (40° + 40°)

= 180° − 80°

= 100°

We show that, AB = CD

∴ ∠COD = ∠AOB [Equal chords subtends equal angle at the center]

= 100°

Thus, the measure of the angle subtended by the chord CD and the radius of circle are respectively 100° and 13 cm.