A16.0-inch chord is drawn in a circle whose radius is 10.0 inches. what is the angular size of the minor arc of this chord? what is the length of the arc, to the nearest tenth of an inch?

PART I
Angular size of the minor arc . 
Half of the chord an the radius makes a right angled triangle with the radius as the hypotenuse and half of the chord as one of the shorter side.
Therefore, using trigonometric ratio, sine = opp/hyp
                          sine θ = 8/10  where θ is half the minor angle
                                  θ  = 53.13
Therefore, the angular size of the minor arc will be 53.13 × 2  = 106.26°

PART II
The length of an arc is given by (θ/360 )× 2πr
where θ is the angle subtended by the arc to the center of the circle and r is the radius of the circle.
Therefore, length = (106.26/360) × 3.142 × 2×10
                            = 18.548 Inches