The circumference of the path traveled by the tip of the minute hand over the course of one hour is:

\(\displaystyle C = 2 \pi r = 2 \pi \cdot 4 = 8 \pi\) feet.

Since the tip of the minute hand has traveled \(\displaystyle 10 \pi\) feet since noon, the minute hand has made 

\(\displaystyle \frac{10 \pi }{8 \pi } = \frac{5}{4} = 1 \frac{1}{4}\) revolutions. Therefore, \(\displaystyle 1 \frac{1}{4}\) hours have elapsed since noon, making the time 1:15 PM.

An inscribed angle of a circle that intercepts a semicircle is a right angle; therefore, \(\displaystyle \angle B\), which intercepts the semicircle \(\displaystyle \overarc {ABC}\), is such an angle. Consequently, \(\displaystyle \bigtriangleup ABC\) is a right triangle, and \(\displaystyle \angle A\) and \(\displaystyle \angle C\) are complementary angles. Therefore,

 \(\displaystyle m \angle A + m \angle C = 90 ^{\circ }\)

\(\displaystyle y + \left ( \frac{y}{3}+2\right ) = 90\)

\(\displaystyle y + \frac{1}{3}y +2 = 90\)

\(\displaystyle \frac{4}{3}y +2 = 90\)

\(\displaystyle \frac{4}{3}y +2 - 2 = 90 - 2\)

\(\displaystyle \frac{4}{3}y = 88\)

\(\displaystyle \frac{3}{4} \cdot \frac{4}{3}y = \frac{3}{4} \cdot 88\)

\(\displaystyle y = 66\)

\(\displaystyle m \angle C = \left (\frac{y}{3}+ 2 \right ) ^{\circ }=\left ( \frac{66}{3}+ 2 \right )^{\circ }=( 22 + 2 )^{\circ }= 24^{\circ }\)

Inscribed \(\displaystyle \angle C\) intercepts an arc with twice its angle measure; this arc is \(\displaystyle \overarc {AB}\), so 

\(\displaystyle m \overarc {AB} = 2 \cdot m \angle C = 2 \cdot 24 ^{\circ } = 48 ^{\circ }\).

The major arc corresponding to this minor arc, \(\displaystyle \overarc {BCA}\), has measure

\(\displaystyle m \overarc {BCA} = 360 ^{\circ } - m \overarc {AB} = 360 ^{\circ } - 48^{\circ } = 312^{\circ }\)

An inscribed angle of a circle that intercepts a semicircle is a right angle; therefore, \(\displaystyle \angle B\), which intercepts the semicircle \(\displaystyle \overarc{ABC}\), is such an angle. Consequently, 

\(\displaystyle m \angle B = 90 ^{\circ }\)

\(\displaystyle 4x-18 = 90\)

\(\displaystyle 4x - 18 + 18 = 90 + 18\)

\(\displaystyle 4x = 108\)

\(\displaystyle 4x \div 4 = 108 \div 4\)

\(\displaystyle x = 27\)

\(\displaystyle m \angle C = (x+10)^{\circ } = (27+10)^{\circ }= 37^{\circ }\)

Inscribed \(\displaystyle \angle C\) intercepts an arc with twice its angle measure; this arc is \(\displaystyle \overarc{AB}\), so 

\(\displaystyle m \widehat{AB} = 2 \cdot m \angle C = 2 \cdot 37 ^{\circ } = 74 ^{\circ }\).

Inscribed \(\displaystyle \angle ACB\), which measures \(\displaystyle 72 ^{\circ }\), intercepts an arc with twice its measure. That arc is \(\displaystyle \overarc {AB}\), which consequently has measure 

\(\displaystyle 72 ^{\circ } \times 2 = 144 ^{\circ }\).

This makes \(\displaystyle \overarc {AB}\) an arc which comprises 

\(\displaystyle \frac{144}{360} = \frac{144 \div 72}{360 \div 72} = \frac{2}{5}\)

of the circle. 

The circumference of a circle is \(\displaystyle 2 \pi\) multiplied by its radius, so 

\(\displaystyle C = 2 \pi r = 2\pi \cdot 20 = 40 \pi\).

The length of \(\displaystyle \overarc {AB}\) is \(\displaystyle \frac{2}{5}\) of this, or \(\displaystyle \frac{2}{5} \cdot 40 \pi = 16 \pi\).

The length of  \(\displaystyle \overarc{AB }\) comprises \(\displaystyle \frac{10}{120} = \frac{1}{12}\) of the circumference of the circle. Therefore, its degree measure is  \(\displaystyle \frac{1}{12} \cdot 360 ^{\circ } = 30^{\circ }\). Similarly, The length of  \(\displaystyle \overarc{CD}\) comprises \(\displaystyle \frac{20}{120} = \frac{1}{6}\) of the circumference of the circle. Therefore, its degree measure is  \(\displaystyle \frac{1}{6} \cdot 360 ^{\circ } = 60 ^{\circ }\). 

If two secants are constructed to a circle from an outside point, the degree measure of the angle the secants form is half the difference of those of the arcs intercepted - that is, 

\(\displaystyle t = \frac{1}{2} ( m \overarc {CD} - m \overarc{AB})\)

\(\displaystyle t = \frac{1}{2} ( 60 - 30 ) = \frac{1}{2} \cdot 30 = 15\).

The circumference of the circle is the sum of the two arc lengths:

\(\displaystyle 80 + 160 = 240\)

The length of  \(\displaystyle \overarc{AB }\) comprises \(\displaystyle \frac{80}{240} = \frac{1}{3}\) of the circumference of the circle. Therefore, its degree measure is  \(\displaystyle \frac{1}{3} \cdot 360 ^{\circ } = 120 ^{\circ }\). Consequently, \(\displaystyle \overarc{ACB }\) is an arc of degree measure \(\displaystyle 360 ^{\circ } - 120 ^{\circ } = 240^{\circ }\). 

The segments shown are both tangents from \(\displaystyle N\) to the circle. Consequently, the degree measure of the angle they form is half the difference of the angle measures of the arcs they intercept - that is,

\(\displaystyle t = \frac{1}{2} ( m \overarc {ACB} - m \overarc{AB})\)

\(\displaystyle t = \frac{1}{2} ( 240 ^{\circ } - 120^{\circ }) = \frac{1}{2} \cdot 120^{\circ } = 60 ^{\circ }\)

The length of  \(\displaystyle \overarc{AB }\) comprises \(\displaystyle \frac{20}{100} = \frac{1}{5}\) of the circumference of the circle. Therefore, its degree measure is  \(\displaystyle \frac{1}{5} \cdot 360 ^{\circ } = 72 ^{\circ }\). Similarly, The length of  \(\displaystyle \overarc{CD}\) comprises \(\displaystyle \frac{15}{100} = \frac{3}{20}\) of the circumference of the circle. Therefore, its degree measure is  \(\displaystyle \frac{3}{20} \cdot 360 ^{\circ } = 54 ^{\circ }\). 

If two chords cut each other inside the circle, as \(\displaystyle \overline{AC }\) and \(\displaystyle \overline{BD}\) do, and one pair of vertical angles are examined, then the degree measure of each angle is half the sum of those of the arcs intercepted - that is, 

\(\displaystyle t = \frac{1}{2} ( m \overarc{AB}+ m \overarc {CD})\)

\(\displaystyle t = \frac{1}{2} ( 72 ^{\circ } + 54 ^{\circ })\)

\(\displaystyle t = \frac{1}{2} ( 126^{\circ })\)

\(\displaystyle t =63 ^{\circ }\)