Since the clock is a circle, you can determine that the total number of degrees inside the circle is 360. Since a clock has 12 numbers, we can divide 360 by 12 to see what the angle is between two numbers that are right next to each other. Thus, we can see that the angle between two numbers right next to each other is $\displaystyle 30^{\circ}$. However, the clock is reading 5:00, so there are five numbers we have to take in to account. Therefore, we multiply 30 by 5, which gives us $\displaystyle 150^{\circ}$ as our answer.

First, remember that the number of degrees in a circle is 360. Then, figure out how many degrees are in between each number on the face of the clock. Since there are 12 numbers, there are $\displaystyle 30^{\circ}$ between each number. Since the time reads 5:00, multiply $\displaystyle 30\cdot 5$, which yields $\displaystyle 150^{\circ}$.

If two chords intersect inside a circle, both chords are cut in a way such that the products of the lengths of the two chords formed in each are the same - in other words, 

$\displaystyle a \cdot b = 1.5 \cdot 1.5$

or

$\displaystyle ab = 2.25$

Therefore, $\displaystyle ab < 3$.

$\displaystyle \overline{OT}$ is a radius of the circle from the center to the point of tangency of $\displaystyle \overline{NT}$, so 

$\displaystyle \overline{OT} \perp \overline{NT}$,

and $\displaystyle \bigtriangleup NTO$ is a right triangle. The length of leg $\displaystyle \overline{NT}$ is known to be 24. The other leg $\displaystyle \overline{OT}$ is a radius radius; we can find its length by dividing the circumference by $\displaystyle 2 \pi$:

$\displaystyle OT = r = \frac{C}{2\pi} = \frac{14 \pi}{2\pi} = 7$

The length hypotenuse, $\displaystyle \overline{NO}$, can be found by applying the Pythagorean Theorem:

$\displaystyle NO = \sqrt{(NT)^{2} + (OT)^{2}}= \sqrt{24^{2} + 7^{2}}=\sqrt{576 + 49} = \sqrt{625} = 25$.

If two chords intersect inside a circle, both chords are cut in a way such that the products of the lengths of the two chords formed in each are the same - in other words, 

$\displaystyle t \cdot 4t = 8 \cdot 24$

Solving for $\displaystyle t$:

$\displaystyle 4t ^{2} = 192$

$\displaystyle 4t ^{2} \div 4 = 192 \div 4$

$\displaystyle t ^{2} = 48$

Since $\displaystyle t ^{2} < 49$, it follows that $\displaystyle t < \sqrt{49 }$, or $\displaystyle t < 7$.

If a secant segment and a tangent segment are constructed to a circle from a point outside it, the square of the distance to the circle along the tangent is equal to the product of the distances to the two points on the circle along the secant; in other words,

$\displaystyle NA \cdot NB = (NT )^{2}$

$\displaystyle NA \cdot (NA + AB)= (NT )^{2}$

$\displaystyle t \cdot (t+t) = 16 ^{2}$

Simplifying, then solving for $\displaystyle t$:

$\displaystyle t \cdot (2t) = 256$

$\displaystyle 2t ^{2} = 256$

$\displaystyle 2t ^{2} \div 2 = 256 \div 2$

$\displaystyle t ^{2} = 128$

To compare $\displaystyle NB$ to 32, it suffices to compare their squares: 

$\displaystyle NB = NA + AB= t + t = 2t$, so, applying the Power of a Product Principle, then substituting,

$\displaystyle (NB)^{2} = (2t) ^{2} = 2^{2} \cdot t^{2} = 4 t ^{2} = 4 \cdot 128 = 512$

$\displaystyle 32^{2} = 1,024$

$\displaystyle 512 < 1,024$, so

$\displaystyle (NB)^{2} < 32 ^{2}$; 

it follows that

$\displaystyle NB < 32$.

If a secant segment and a tangent segment are constructed to a circle from a point outside it, the square of the distance to the circle along the tangent is equal to the product of the distances to the two points on the circle intersected by the secant; in other words,

$\displaystyle (NT)^{2}= NA \cdot NB$

$\displaystyle (NT)^{2}= NA \cdot (NA+AB)$

$\displaystyle (2t) ^{2} = t (t + 24)$

Simplifying and solving for $\displaystyle t$:

$\displaystyle 2^{2} \cdot t^{2} = t \cdot t + t \cdot 24$

$\displaystyle 4 t^{2} = t ^{2} + 24t$

$\displaystyle 4 t^{2} - t ^{2} - 24t = t ^{2} + 24t - t ^{2} - 24t$

$\displaystyle 3 t^{2}- 24t = 0$

Factoring out $\displaystyle 3t$:

$\displaystyle 3 t (t-8) = 0$

Either $\displaystyle t = 0$ - which is impossible, since $\displaystyle t$ must be positive, or

$\displaystyle t- 8 =0$, in which case $\displaystyle t = 8$.

If two chords intersect inside a circle, both chords are cut in a way such that the products of the lengths of the two chords formed in each are the same - in other words, 

$\displaystyle ab = cd$

Divide both sides of this equation by $\displaystyle bd$, then cancelling:

$\displaystyle \frac{ab }{bd}= \frac{cd}{bd}$

$\displaystyle \frac{a }{d}= \frac{c}{b}$

The two quantities are equal.

Let $\displaystyle r$ be the radius of Circle A. Then its area is $\displaystyle \pi r^{2}$.

The area of Circle B is $\displaystyle 4\pi r^{2} =2^{2}\pi r^{2} = \pi (2r)^{2}$, so the radius of Circle B is twice that of Circle A; by a similar argument, the radius of Circle C is twice that of Circle B, or $\displaystyle 2 (2r ) = 4r$.

(a) Twice the radius of circle B is $\displaystyle 2 (2r) = 4r$.

(b) The sum of the radii of Circles A and B is $\displaystyle r + 4r = 5r$.

This makes (b) greater.

Every hour, the tip of the minute hand travels the circumference of a circle. Between noon and 1:45 PM, one and three-fourths hours pass, so the tip travels $\displaystyle 1 \frac{3}{4}$ or $\displaystyle \frac{7}{4}$ times this circumference. The length of the minute hand is the radius of this circle $\displaystyle r$, and the circumference of the circle is $\displaystyle C = 2 \pi r$, so the distance the tip travels is $\displaystyle \frac{7}{4}$ this, or

$\displaystyle \frac{7}{4} C = \frac{7}{4} \cdot 2 \pi r = \frac{7 \pi r }{2}$

Set this equal to $\displaystyle \frac{14\pi}{3}$ feet:

$\displaystyle \frac{7 \pi r }{2} = \frac{14\pi}{3}$

$\displaystyle \frac{7 \pi r }{2} \times \frac{2}{7 \pi}= \frac{14\pi}{3}\times \frac{2}{7 \pi}$

$\displaystyle r= \frac{2}{3}\times \frac{2}{1} = \frac{4}{3} = 1 \frac{1}{3}$ feet.

This is equivalent to 1 foot 4 inches.