This is simply a matter of percentages. We first have to figure out what percentage of the surface area is represented by 4.5π. To do that, we must calculate the total surface area. If the diameter is 12, the radius is 6. Don't be tricked by this!

A = π * 6 * 6 = 36π

Now, 4.5π is 4.5π/36π percentage or 0.125 (= 12.5%)

To figure out the angle, we must take that percentage of 360°:

0.125 * 360 = 45°

12 compartments further means 240 more degrees. 240 is the answer.

360/12 = 240 degrees

The measure of angle CEF is going to be equal to half of the difference between the measures two arcs that it intercepts, namely arcs AD and CD.

$\displaystyle \angle CEF=\frac{1}{2}(AD-CD)$

Thus, we need to find the measure of arcs AD and CD. Let's look at the information given and determine how it can help us figure out the measures of arcs AD and CD. 

Angle BAC is an inscribed angle, which means that its meausre is one-half of the measure of the arc that it incercepts, which is arc BC.

$\displaystyle \angle BAC=\frac{1}{2}BC$

$\displaystyle 35^o=\frac{1}{2}BC\rightarrow BC=70^o$

Thus, since angle BAC is 35 degrees, the measure of arc BC must be 70 degrees.

We can use a similar strategy to find the measure of arc CD, which is the arc intercepted by the inscribed angle FBD.

$\displaystyle 40^o=\frac{1}{2}CD\rightarrow CD=80^o$

Because angle FBD has a measure of 40 degrees, the measure of arc CD must be 80 degrees.

We have the measures of arcs BC and CD. But we still need the measure of arc AD. We can use the last piece of information given, along with our knowledge about the sum of the arcs of a circle, to determine the measure of arc AD.

We are told that the measure of arc AD is twice the measure of arc AB. We also know that the sum of the measures of arcs AD, AB, CD, and BC must be 360 degrees, because there are 360 degrees in a full circle. 

$\displaystyle AD+AB+CD+BC=360^o$

$\displaystyle AD+AB+80^o+70^o=360^o$

$\displaystyle AD+AB=210^o$

Because AD = 2AB, we can substitute 2AB for AD.

$\displaystyle 2AB+AB=210^o$

$\displaystyle 3AB=210^o$

$\displaystyle AB=70^o$

This means the measure of arc AB is 70 degrees, and the measure of arc AD is 2(70) = 140 degrees.

Now, we have all the information we need to find the measure of angle CEF, which is equal to half the difference between the measure of arcs AD and CD.

$\displaystyle \angle CEF=\frac{1}{2}(AD-CD)$

$\displaystyle \angle CEF=\frac{1}{2}(140^o-80^o)=\frac{1}{2}(60^o)$

$\displaystyle \angle CEF=30^o$

The circumference of the circle is $\displaystyle 2\pi r$.

$\displaystyle 2\pi(16)=32\pi$

The length of the arc S is $\displaystyle 8\pi$.

A ratio can be established:

$\displaystyle \frac{S}{\text{circumference}}=\frac{\theta}{360^o}$

$\displaystyle \frac{8\pi}{32\pi}=\frac{\theta}{360^o}$

Solving for $\displaystyle \theta$yields 90^o. 

Note: This makes sense. Since the arc S was one-fourth the circumference of the circle, the central angle formed by arc S should be one-fourth the total degrees of a circle.

Since we know that segment AC is a diameter, this means that the length of the arc ABC must be 180 degrees. This means that the length of the arc AB must be 80 degrees. 

Since angle ADB is an inscribed angle, its measure is equal to half of the measure of the angle of the arc that it intercepts. This means that the measure of the angle is half of 80 degrees, or 40 degrees.

To begin, you should compute the complete area of the circle:

$\displaystyle A=\pi r^2$

For your data, this is:

$\displaystyle A=15^2\pi=225\pi$

Now, to find the angle measure of a sector, you find what portion of the circle the sector is. Here, it is:

$\displaystyle \frac{45}{225\pi}$

Now, multiply this by the total $\displaystyle 360$ degrees in a circle:

$\displaystyle \frac{45}{225\pi}*360=22.918311805212$

Rounded, this is $\displaystyle 22.92^{\circ}$.

The first thing to do for this problem is to compute the total circumference of the circle. Notice that you were given the diameter. The proper equation is therefore:

$\displaystyle C=\pi d$

For your data, this means,

$\displaystyle C=12\pi$

Now, to compute the angle, note that you have a percentage of the total circumference, based upon your arc length:

$\displaystyle \frac{13.5}{12\pi}*360=128.9155039044336$

Rounded to the nearest hundredth, this is $\displaystyle 128.92^{\circ}$.

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, $\displaystyle m \angle B = 90 ^{\circ }$, and $\displaystyle \bigtriangleup ABC$ is a right triangle. The acute angles of $\displaystyle \bigtriangleup ABC$ are complementary, so

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

$\displaystyle (7t+1 )+ (4t+1) = 90$

$\displaystyle 7t+4t+1+1 = 90$

$\displaystyle 11t+2 = 90$

$\displaystyle 11t+2 -2 = 90 - 2$

$\displaystyle 11t = 88$

$\displaystyle 11t\div 11 = 88 \div 11$

$\displaystyle t = 8$

The measure of inscribed $\displaystyle \angle C$ is 

$\displaystyle m \angle C = (4t+1 )^{\circ } = (4 \cdot 8+1 )^{\circ } = 33 ^{\circ }$.

An inscribed angle of a circle intercepts an arc of twice its degree measure, so

$\displaystyle m \overarc {AB} = 2 \cdot m \angle C= 2 \cdot 33^{\circ } = 66 ^{\circ }$.

$\displaystyle \overline{AB}$ is a diameter, so $\displaystyle \overarc{BC A }$ is a semicircle, and

$\displaystyle m \overarc{BC A } = 180^{\circ }$,

or, equivalently,

 $\displaystyle m \overarc{BC} + m \overarc {AC } = 180^{\circ }$

In terms of $\displaystyle t$, since $\displaystyle m \overarc {AC } = t^{\circ }$,

$\displaystyle m \overarc{BC} + t ^{\circ }= 180^{\circ }$

$\displaystyle m \overarc{BC} + t^{\circ } - t ^{\circ }= 180 ^{\circ } - t^{\circ }$

$\displaystyle m \overarc{BC} =( 180 - t)^{\circ }$

$\displaystyle \overline{NB}$ and $\displaystyle \overline{NC }$, being a secant segment and a tangent segment to a circle, respectively, intercept two arcs such that the measure of the angle that the segments form is equal to one-half the difference of the measures of the intercepted arcs - that is,

$\displaystyle \frac{1}{2} (m \overarc{BC} - m \overarc {AC }) = m \angle CNB$

Setting $\displaystyle m \overarc{BC} = \left (180 - t \right )^{\circ }$, $\displaystyle m \overarc {AC } = t^{\circ }$, and $\displaystyle m \angle CNB = 34^{\circ }$:

$\displaystyle \frac{1}{2}\left [ (180 - t) - t \right ]= 34$

$\displaystyle \frac{1}{2} (180 - 2t) = 34$

$\displaystyle \frac{1}{2} \cdot 180 -\frac{1}{2} \cdot 2t = 34$

$\displaystyle 90 - t= 34$

$\displaystyle 90-t-90 = 34 -90$

$\displaystyle -t =- 56$

$\displaystyle t = 56$

The total measure of the arcs that comprise a circle is $\displaystyle 360 ^{\circ }$, so from the above diagram,

$\displaystyle m \overarc { AC}+m \overarc { CD}+ m \overarc { DB} + m \overarc {BA } = 360 ^{\circ }$

Substituting the appropriate expression for each arc measure:

$\displaystyle t+t+3t+3t = 360$

$\displaystyle 8t = 360$

$\displaystyle 8t \div 8 = 360 \div 8$

$\displaystyle t = 45$

Therefore, 

$\displaystyle m \overarc { AC}= 45^{\circ }$ 

and 

$\displaystyle m \overarc { DB} = 3 t^{\circ } = 3 \cdot 45^{\circ } = 135^{\circ }$

The measure of the angle formed by the tangent segments $\displaystyle \overline{NB}$ and $\displaystyle \overline{ND}$, which is $\displaystyle \angle BND$, is half the difference of the measures of the arcs they intercept, so 

$\displaystyle m\angle BND = \frac{1}{2} (m \overarc {DB} - m \overarc {AC})$

Substituting:

$\displaystyle m\angle BND = \frac{1}{2} (135 ^{\circ }- 45^{\circ } )$

$\displaystyle m\angle BND = \frac{1}{2} (90^{\circ } )$

$\displaystyle m\angle BND = 45 ^{\circ }$