Every hour, the tip of the minute hand travels the circumference of a circle with radius six feet, which is

\(\displaystyle C = 2 \pi r= 2 \pi \cdot 6 = 12 \pi\) feet.

Since it is now 3:50 PM, the minute hand made three complete revolutions since noon, plus \(\displaystyle \frac{50}{60 } = \frac{5}{6}\) of a fourth, so its tip has traveled this circumference \(\displaystyle 3\frac{5}{6}\) times. 

This is

\(\displaystyle 3\frac{5}{6} \times 12 \pi = \frac{23}{6} \times 12 \pi = 46 \pi\) feet. This is 

\(\displaystyle 46 \pi \times 12 = 552 \pi\) inches.

The tip of a minute hand travels a circle whose radius is equal to the length of that minute hand, which, in this question, is seven feet long. The circumference of this circle is \(\displaystyle 2 \pi\) times the radius, or \(\displaystyle 2 \pi \times 7 = 14\pi\) feet; over the course of thrity minutes (or one-half of an hour) the tip of the minute hand covers half this distance, or \(\displaystyle \frac{1}{2} \times 14 \pi = 7 \pi\) feet.

The circumference of a circle seven feet in diameter is \(\displaystyle \pi\) times this diameter, or \(\displaystyle 7 \pi\) feet.

The quantities are equal.

Every hour, the tip of the minute hand travels the circumference of a circle, which here is 

\(\displaystyle C = 2 \pi r = 2 \pi \times 4 \frac{1}{2} = 9 \pi\) feet.

The minute hand has traveled \(\displaystyle 110 \pi\) feet since noon, so it has traveled the circumference of the circle

\(\displaystyle \frac{110 \pi}{9 \pi} = \frac{110 }{9 } =12 \frac{2}{9}\) times.

Since \(\displaystyle 12 \leq 12 \frac{2}{9} \leq 12 \frac{1}{2}\), between 12 and \(\displaystyle 12 \frac{1}{2}\) hours have elapsed since noon, and the time is between 12:00 midnight and 12:30 AM.

Examine the figure below, which shows \(\displaystyle \bigtriangleup ABC\) inscribed in a circle.

By the Arc Addition Principle,

\(\displaystyle m \overarc {ABC} = m \overarc {AB}+ m \overarc {BC}\)

and

\(\displaystyle m \overarc {ACB} = m \overarc {AC}+ m \overarc {BC}\)

Consequently,

\(\displaystyle m \overarc {ABC} - m \overarc {ACB}\)

\(\displaystyle = (m \overarc {AB}+ m \overarc {BC} ) - ( m \overarc {AC}+ m \overarc {BC})\)

\(\displaystyle = m \overarc {AB}+ m \overarc {BC} - m \overarc {AC}- m \overarc {BC}\)

\(\displaystyle = m \overarc {A B} - m \overarc {A C}\)

The two quantities are equal. 

(a) A circle with radius 8 inches has crircumference \(\displaystyle C = 2\pi r = 2\pi \cdot 8 = 16\pi\) inches. An arc 10 inches long is \(\displaystyle \frac{10}{16\pi } = \frac{5}{8\pi }\) of that circle. \(\displaystyle \frac{5}{8\pi } \times 360 ^{\circ } = \frac{225}{\pi } ^{\circ }\), the degree measure of this arc.

(b) A circle with radius 10 inches has crircumference \(\displaystyle C = 2\pi r = 2\pi \cdot 10 = 20\pi\) inches. An arc 12 inches long is \(\displaystyle \frac{12}{20\pi }=\frac{3}{5\pi }\) of that circle. \(\displaystyle \frac{3}{5\pi } \times 360 ^{\circ } = \frac{216}{\pi } ^{\circ }\), the degree measure of this arc.

(a) is the greater quantity.

Since 

\(\displaystyle 7^{2} + 12 ^{2} = 49 + 144 = 169 = 13 ^{2}\),

the triangle is a right triangle with right angle \(\displaystyle \angle ACB\).

\(\displaystyle \angle ACB\) is an inscribed angle on the circle, so the arc it intercepts is a semicricle. Therefore, \(\displaystyle \widehat{ACB}\) is also a semicircle, and it measures \(\displaystyle 180 ^{\circ }\).

The measure of an arc intercepted by an inscribed angle of a circle is twice that of the angle. Therefore, \(\displaystyle y = 2 \cdot 44 = 88 < 90\)

\(\displaystyle \odot A\) has twice the radius of \(\displaystyle \odot B\), so \(\displaystyle \odot A\) has four times the area of \(\displaystyle \odot B\). This means that for a sector of \(\displaystyle \odot A\) to have the same area as a sector of \(\displaystyle \odot B\), the central angle of the latter sector must be four times that of the former sector. This makes (b) greater than (a), which is only twice that of the former sector.

The measure of an inscribed angle of a circle is one-half that of the arc it intercepts. Therefore, \(\displaystyle x = \frac{1}{2} \cdot 110 = 55\).

Remember that the angle for a sector or arc is found as a percentage of the total \(\displaystyle 360\) degrees of the circle.  The proportion of \(\displaystyle x\) to \(\displaystyle 360\) is the same as \(\displaystyle 19.4\) to the total circumference of the circle.

The circumference of a circle is found by:

\(\displaystyle C = 2*\pi*r\)

For our data, this means:

\(\displaystyle C = 2*8.7*\pi=17.4\pi\)

Now we can solve for \(\displaystyle x\) using the proportions:

\(\displaystyle \frac{x}{360} = \frac{19.4}{17.4\pi}\)

Cross multiply:

\(\displaystyle 17.4x\pi=6984\)

Divide both sides by \(\displaystyle 17.4\pi\):

\(\displaystyle x=127.76300259239047\)

Therefore, \(\displaystyle x\) is \(\displaystyle 127.76\)˚.